Iteration series
#2
I'll explain the use of iteration-series in more detail.

The formula for the binomial-method, (which Henryk announced to the forum last year, was shown by Woon, and earlier by Comtet who credits again earlier use by Jabotinsky) is
\( \hspace{24}
f^{\circ h}(x) = \sum_{k=0}^{\infty}\left( (^h_k) \left( \sum_{j=0}^k (-1)^{k-j} (^k_j) f^{\circ j}(x) \right) \right)
\)

The most straigthforward translation of my application of the matrix-method into the similar serial notation is perhaps the following. I'm writing \( \hspace{24} [x]_0^m \) for the pochhammer-symbol x(x-1)(x-2)...(x-m)

derived from the logarithmic-formula exp(h*log(...)) using the matrices of Stirling-numbers I got
\( \hspace{24}
f^{\circ h}(x) = 1 + \sum_{k=0}^{\infty} \left(
\left(\sum_{j=0}^k (-1)^{k-j} (^k_j) f^{\circ j}(x) \right)
* [h]_0^{k-1} /k!
\right)
\)

[update] Upps... second read: seems I just reduced the combined exp( h* log() ) formulae to arrive at the same as above Rolleyes But well, I'll leave it as it is, the main focus is the following) [/update]


Well, it's not my intention at the moment to compare approximation-rates and convergence issues. It is just to introduce the translation from the concept in my matrix-formula (which is taylored for algebra with formal powerseries) to the equivalent formulation for iteration-series, where only integer-iteration-heights are required to arrive at (and possibly further use) fractional heights.

The whole concept is then much more easier to read (and write) if we introduce an "umbral"-like notation, adapted to iteration-series.

So my proposal tonight is,to write \( \left[f^{\circ}(x) + I^{\circ}\right]^m \) meaning the binomial-formula, in which after expansion the powers are understood as heights. (I is here the identity-operator.)

So \( \hspace{24} [f^{\circ}(x) +I^{\circ}]^3 \) expands formally to

\( \hspace{24}
[f^{\circ}(x) +I^{\circ}]^3 = f^{\circ 3}(x)I^{\circ 0} +3 f^{\circ 2}(x)I^{\circ 1}
+ 3 f^{\circ 1}(x)I^{\circ 2} + f^{\circ 0}(x)I^{\circ 3} \\
\hspace{48}
= f^{\circ 0}(x) +3 f^{\circ 1}(x) + 3^{\circ 2}(x) + f^{\circ 3}(x)
\)

and the analogue interpretation may be understood by the notation \( \log\left[f^{\circ}(x)\right] \),\( \exp\left[f^{\circ}(x)\right] \) for log(), exp() and other functions, which are expressed by powerseries.

Let us also omit the "(x)" at the function, to make things shorter.
Then with this newly introduced Iteration-Umbral we express the above binomial-formula

\( \hspace{24}
f^{\circ h} = [I^{\circ}+ (f^{\circ}-I^{\circ})^{\circ}]^h \\
\hspace{48} \left( = \sum_{k=0}^{\infty}\left( (^h_k) (f^{\circ}-I^{\circ})^k \right) \right) \\
\hspace{48} \left( = \sum_{k=0}^{\infty}\left( (^h_k) \left( \sum_{j=0}^k (-1)^{k-j} (^k_j) f^{\circ j} \right) \right) \right)
\)

and the logarithmic formula

\( \hspace{24}
f^{\circ h} = \exp\left[h * \log\left[I^{\circ}+ (f^{\circ}-I^{\circ})^{\circ}\right] \right]
\)

May be, the short-notation can further be polished, for instance first I thought to use an additional placeholder at the formal tiny iteration-circle - maybe such expliciteness is sometimes required in more complicated places.

Gottfried
Gottfried Helms, Kassel
Reply


Messages In This Thread
Iteration series - by Gottfried - 02/22/2009, 10:20 PM
RE: Iteration series - by Gottfried - 02/26/2009, 05:18 AM
RE: Iteration series - by bo198214 - 02/28/2009, 10:02 AM
RE: Iteration series - by Gottfried - 08/10/2009, 04:16 PM

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