While I played around with some transformations to find a meaningful transformation for a summation in the divergent case, I just found a transformation of the problem, which at least provides an improvement for the approximation for the convergent case.
Henryk's formula is (I inserted (x) at f°t)
\( \hspace{24} f^{\circ t}(x) = \sum_{n=0}^{\infty} ({t\\n})* \sum_{j=0}^n * (-1)^{n+j}* ({n\\j})* f^{\circ j}(x)) \)
Here I rewrite this in vectorial notation, for simpliness.
With this Henryk's formula may be rewritten as product of vectors V1 and V2
\( \hspace{24}
f^{\circ t}(x) = V_1\sim * V_2 \)
where
\( \hspace{24}
V_1 = \text{col}_{\tiny r=0}^{\tiny \infty} [({t\\r})] \)
\( \hspace{24}
V_2 = \text{col}_{\tiny r=0}^{\tiny \infty}[ \sum_{c=0}^r (-1)^{r+c}*({r\\c})*f^{\circ c}(x) ) ] \)
("col" means here: a column-vector of index/length as subscripted)
into
\( \hspace{24}
f^{\circ t}(x) = V_1\sim * V_2 \)
where
\( \hspace{24}
V_1 = \text{col}_{\tiny r=0}^{\tiny \infty}[ \sum_{c=0}^r ((-1)^{r+c}* ({r\\c})*c^t ) ]
\)
\( \hspace{24}
V_2 = \text{col}_{\tiny r=0}^{\tiny \infty}[ \sum_{c=0}^r ( \frac{S1(r,c)}{r!} * f^{\circ c}(x) ) ]
\)
Here S1(r,c) means the Stirlingnumber 1st kind of row=r,col=c
Using f(x) = b^x, f°t(x) = f°(t-1)(b^x) , with b=sqrt(2) and t=0.5 I get different series with the two methods.
Here is the comparison (the numbers are the individual terms from the vectorproduct V1~*V2):
\( \hspace{24}
\begin{matrix} {ll}
\text{binomial} & \text{stirling} \\
1 & . \\
0.207106781187 & 1.41421356237 \\
0.0244875256635 & -0.0639425018608 \\
0.00661871779280 & -0.0251486716496 \\
0.00250547215205 & -0.0138115634657 \\
0.00115807690080 & -0.00888364117548 \\
0.000610766543495 & -0.00626721909452 \\
0.000353479520962 & -0.00469791870292 \\
0.000219183130875 & -0.00367580993247 \\
0.000143364426110 & -0.00296926498063 \\
0.0000978609982638 & -0.00245835601888 \\
0.0000691742158295 & -0.00207564611019 \\
0.0000503406857704 & -0.00178071233011 \\
0.0000375480921629 & -0.00154805648671 \\
0.0000286035266271 & -0.00136090991516 \\
0.0000221916461008 & -0.00120785502708 \\
0.0000174945666035 & -0.00108088575828 \\
0.0000139875666870 & -0.000974244878765 \\
\ldots \text{ } & \ldots \\
\ldots \text{ continue} & \text{at row 90}\ldots \\
\ldots \text{ } & \ldots \\
0.0000000285776133075 & -0.0000638676539217 \\
0.0000000274286147892 & -0.0000627660387118 \\
0.0000000263376884568 & -0.0000616955986773 \\
0.0000000253013130350 & -0.0000606551185701 \\
0.0000000243162155134 & -0.0000596434431489 \\
0.0000000233793512466 & -0.0000586594736066 \\
\ldots & \ldots \\
\end{matrix}
\)
The problem for the approximation of a useful result based on these methods is the (slow) monotone decrease (or increase) of these series and hence of the partial sums. That may also be the reason for the observed difference of the diagonalization-method to this type of computation: even using 200 and more terms (as I did) leaves a remainder in the order of 1e-10 or the like.
The nice thing is now, that we have an interval for the result, from where we may interpolate a better estimate. However, I don't see it yet (to use the mean seems to be too simple...)
Henryk's formula is (I inserted (x) at f°t)
\( \hspace{24} f^{\circ t}(x) = \sum_{n=0}^{\infty} ({t\\n})* \sum_{j=0}^n * (-1)^{n+j}* ({n\\j})* f^{\circ j}(x)) \)
Here I rewrite this in vectorial notation, for simpliness.
With this Henryk's formula may be rewritten as product of vectors V1 and V2
\( \hspace{24}
f^{\circ t}(x) = V_1\sim * V_2 \)
where
\( \hspace{24}
V_1 = \text{col}_{\tiny r=0}^{\tiny \infty} [({t\\r})] \)
\( \hspace{24}
V_2 = \text{col}_{\tiny r=0}^{\tiny \infty}[ \sum_{c=0}^r (-1)^{r+c}*({r\\c})*f^{\circ c}(x) ) ] \)
("col" means here: a column-vector of index/length as subscripted)
into
\( \hspace{24}
f^{\circ t}(x) = V_1\sim * V_2 \)
where
\( \hspace{24}
V_1 = \text{col}_{\tiny r=0}^{\tiny \infty}[ \sum_{c=0}^r ((-1)^{r+c}* ({r\\c})*c^t ) ]
\)
\( \hspace{24}
V_2 = \text{col}_{\tiny r=0}^{\tiny \infty}[ \sum_{c=0}^r ( \frac{S1(r,c)}{r!} * f^{\circ c}(x) ) ]
\)
Here S1(r,c) means the Stirlingnumber 1st kind of row=r,col=c
Using f(x) = b^x, f°t(x) = f°(t-1)(b^x) , with b=sqrt(2) and t=0.5 I get different series with the two methods.
Here is the comparison (the numbers are the individual terms from the vectorproduct V1~*V2):
\( \hspace{24}
\begin{matrix} {ll}
\text{binomial} & \text{stirling} \\
1 & . \\
0.207106781187 & 1.41421356237 \\
0.0244875256635 & -0.0639425018608 \\
0.00661871779280 & -0.0251486716496 \\
0.00250547215205 & -0.0138115634657 \\
0.00115807690080 & -0.00888364117548 \\
0.000610766543495 & -0.00626721909452 \\
0.000353479520962 & -0.00469791870292 \\
0.000219183130875 & -0.00367580993247 \\
0.000143364426110 & -0.00296926498063 \\
0.0000978609982638 & -0.00245835601888 \\
0.0000691742158295 & -0.00207564611019 \\
0.0000503406857704 & -0.00178071233011 \\
0.0000375480921629 & -0.00154805648671 \\
0.0000286035266271 & -0.00136090991516 \\
0.0000221916461008 & -0.00120785502708 \\
0.0000174945666035 & -0.00108088575828 \\
0.0000139875666870 & -0.000974244878765 \\
\ldots \text{ } & \ldots \\
\ldots \text{ continue} & \text{at row 90}\ldots \\
\ldots \text{ } & \ldots \\
0.0000000285776133075 & -0.0000638676539217 \\
0.0000000274286147892 & -0.0000627660387118 \\
0.0000000263376884568 & -0.0000616955986773 \\
0.0000000253013130350 & -0.0000606551185701 \\
0.0000000243162155134 & -0.0000596434431489 \\
0.0000000233793512466 & -0.0000586594736066 \\
\ldots & \ldots \\
\end{matrix}
\)
The problem for the approximation of a useful result based on these methods is the (slow) monotone decrease (or increase) of these series and hence of the partial sums. That may also be the reason for the observed difference of the diagonalization-method to this type of computation: even using 200 and more terms (as I did) leaves a remainder in the order of 1e-10 or the like.
The nice thing is now, that we have an interval for the result, from where we may interpolate a better estimate. However, I don't see it yet (to use the mean seems to be too simple...)
Gottfried Helms, Kassel