Hmm, one more comment, why the property \( f°^{a+b}(z) = f°^b(f°^a(z)) \) holds for \( f(z) = \sin(z) \)
The Carleman-ansatz gives a triangular Carlemanmatrix, say \( S \) for the sine-function. \( S \) however has the diagonal of \( 1 \), so the diagonalization (which were then the operationalizing of the Schroeder-mechanism) cannot be applied here.
But for any finite size of \( S \) we can determine exactly the matrix logarithm using the Mercator-series for the \( \log(1+x) \) applied to the matrix \( S - I \). We get then -say-
\[ L = \text{matlog}(S) \tag 1\]
(Note: we generalize this to the case of infinite size because the entries in the matrix do not change if we increase size; this is due to triangularity of and the unit diagonal in \( S \) )
Equivalently we can apply the powerseries for the exponential-function to \( L \) getting exact, rational, coefficients.
We can thus formulate for some (integer or fractional) iteration-height \( h \)
\[ S^h = \text{matexp}( h \cdot L) \tag 2
\]
In Pari/GP the parameter \( h \) in this operations can be left symbolic so we get exact rational expressions in \( S^h \) - and the relevant coefficients \( s_{r,h} \) are in the second column and r'th row of \( S^h \) . They are exactly the polynomials in \(h \) as shown by Daniel (and are also well known elsewhere).
By this it is obvious that
\[ S^a \cdot S^b = \text{matexp}( b \cdot L) \cdot \text{matexp}( a \cdot L) = \text{matexp}( ( a + b) \cdot L) = S^{a+b} \tag 3
\]
down to the level of the coefficients of the powerseries-to-be-generated by the common rule of the exponential \( e^{\lambda a} \cdot e^{\lambda b} = e^{\lambda (a+b)} \).
- - - - - - - -
Whether the so-found powerseries \[ \sin°^h(z) = s_{1,h} \cdot z + s_{2,h} \cdot z^2 + s_{3,h} \cdot z^3 + ... \tag 4 \] can be evaluated for fractional values \(h \) and \( z \) other than zero is another question and needs then (strong) procedures of divergent summation (I have applied selfmade such procedures and have approximated examplaric values for instance for \( h =0.5 \)) . Perhaps (or "likely"?) there are more sophisticated procedures for the evaluation of that powerseries around, but I didn't search for that so far.
See also my answer1 and/or answer2 in MO where this is a bit described and in which complete sequence of answers are many more valuable informations.
Gottfried
The Carleman-ansatz gives a triangular Carlemanmatrix, say \( S \) for the sine-function. \( S \) however has the diagonal of \( 1 \), so the diagonalization (which were then the operationalizing of the Schroeder-mechanism) cannot be applied here.
But for any finite size of \( S \) we can determine exactly the matrix logarithm using the Mercator-series for the \( \log(1+x) \) applied to the matrix \( S - I \). We get then -say-
\[ L = \text{matlog}(S) \tag 1\]
(Note: we generalize this to the case of infinite size because the entries in the matrix do not change if we increase size; this is due to triangularity of and the unit diagonal in \( S \) )
Equivalently we can apply the powerseries for the exponential-function to \( L \) getting exact, rational, coefficients.
We can thus formulate for some (integer or fractional) iteration-height \( h \)
\[ S^h = \text{matexp}( h \cdot L) \tag 2
\]
In Pari/GP the parameter \( h \) in this operations can be left symbolic so we get exact rational expressions in \( S^h \) - and the relevant coefficients \( s_{r,h} \) are in the second column and r'th row of \( S^h \) . They are exactly the polynomials in \(h \) as shown by Daniel (and are also well known elsewhere).
By this it is obvious that
\[ S^a \cdot S^b = \text{matexp}( b \cdot L) \cdot \text{matexp}( a \cdot L) = \text{matexp}( ( a + b) \cdot L) = S^{a+b} \tag 3
\]
down to the level of the coefficients of the powerseries-to-be-generated by the common rule of the exponential \( e^{\lambda a} \cdot e^{\lambda b} = e^{\lambda (a+b)} \).
- - - - - - - -
Whether the so-found powerseries \[ \sin°^h(z) = s_{1,h} \cdot z + s_{2,h} \cdot z^2 + s_{3,h} \cdot z^3 + ... \tag 4 \] can be evaluated for fractional values \(h \) and \( z \) other than zero is another question and needs then (strong) procedures of divergent summation (I have applied selfmade such procedures and have approximated examplaric values for instance for \( h =0.5 \)) . Perhaps (or "likely"?) there are more sophisticated procedures for the evaluation of that powerseries around, but I didn't search for that so far.
See also my answer1 and/or answer2 in MO where this is a bit described and in which complete sequence of answers are many more valuable informations.
Gottfried
Gottfried Helms, Kassel