Matrix Operator Method

[bo198214]

The symbolic eigensystem-decomposition is enormous resources-consuming and also not yet well documented and verified in a general manner. Here I propose a completely elementary (while still matrix-based) way to arrive at the same result as by eigensystem-analysis.

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As you might recall from my "continuous iteration" article I've found a matrix representation for the coefficients of the powerseries for the U-tetration
\( \hspace{24} U_t^{^oh}(x) \), where \( U_t(x)= U_t^{^o1}(x) = t^x - 1 \).

Gottfried, as long as you deal with functions with fixed point at 0, for example \( b^x-1 \) or \( xb^x \) there is no need to apply the Diagonalization method in its full generality.
Note, that I would like to change the name from matrix operator method to diagonalization method, as this is more specific (for example Walker uses the name "matrix method" for his natural Abel method).

For the case of a fixed point at 0, we can use the ordinary formulas of regular iteration. You know there are two cases, first \( f'(0)=1 \) (which Daniel Geisler calls parabolic) and \( 0<f'(0)<1 \) or \( f'(0)>1 \) (which Daniel Geisler calls hyperbolic, however I am not really happy with those names, because where is the elliptic case? Is this \( 0<f'(0)<1 \) in opposite to the \( f'(0)>1 \) which we then would call "properly hyperbolic". Also the hyperbolic and elliptic cases (in the just mentioned sense) exchange by using \( f^{-1} \) and hence can be treated quite similarly. This is not the case for hyperbolas and ellipses.)

The formula for the parabolic iteration was already mentioned on this forum under the name double binomial formula.

The formula for hyperbolic iteration (or elliptic iteration) can be derived from the fact that
\( f^{\circ t}\circ f=f\circ f^{\circ t} \)
and by the regularity condition:
\( (f^{\circ t})'(0)=(f'(0))^t \)

For abbreviation let \( g:=f^{\circ t} \), then the first equation can be written as
\( g_n (f^n)_n + \sum_{m=1}^{n-1} g_m (f^m)_n = f_1 g_n + \sum_{m=2}^n f_m (g^m)_n \)
where \( f^n \) means the \( n \)th power, by the subscript index \( i \) we denote the \( i \)th coefficients of the indexed powerseries.

We can rearrange the above formula to get a recurrence relation for the coefficients of \( g \):

\( g_n = \frac{1}{(f_1)^n-f_1}\left( f_n (g_1)^n - g_1 f_n +
\sum_{m=2}^{n-1} (f_m (g^m)_n - g_m (f^m)_n) \right) \)

the only undetermined coefficient is now \( g_1 \) but we know already that we set \( g_1=f_1^t \). The coefficient \( (f^m)_n \) is the entry at the m-th row and the n-th column of the Carleman matrix of \( f \). We see that in the above formula also the term \( (g^m)_n \) occurs though we dont know all the coefficients of \( g \) yet, however we have the power formula:
\( \left(f^p\right)_n=\sum_{
p_1+\dots+p_n=p\\
1p_1+\dots+np_n=n\\
p_1,\dots,p_n\ge 0
}
\frac{p!}{p_1!\dots p_n!} (f_1)^{p_1}\dots
(f_{n})^{p_n} \)

and see that in \( (f^p)_n \) the biggest index taken at \( f \) that can occur is \( n \). However \( p_n \), the exponent of \( f_n \), can be bigger than 0 only in the case \( p_n=1 \) (second line below the sum) and all other \( p_i=0 \). This however can only happen for \( p=1 \) (first line below the sum). This case is excluded in \( (g^m)_n \) because \( m\ge 2 \) and so we have indeed a recurrence formula which gives a polynomial in \( g_1=f_1^t \) for \( g_n \).

You see, we not even have to solve a linear equation system for using the diagonalization method on a fixed point at 0.