ahh, stupid....
Of course the basic Chebychev-polynomials are "the-functions-of which-acosh(x)-is-the-Schröderfunction".
Let for instance \( f_2(x) = T_2(x) = -1 + 2x^2 \) then of course the iterates \( f_2^{\circ 2}(x) = T_2(T_2(x)) = 1 -8x^2+8x^4 \) and the iterates to any height h are \( f_2^{\circ h}(x)= T_2^{\circ h}(x) \) which can then be evaluated/interpolated using the cosh/acosh-pair as \( f_2^{\circ h}(x) = \cosh(2^h \operatorname{acosh}(x)) \) .
So my question is then -very simple- answered for "base=2" that the "more common" function "in our toolbox" is the quadratic polynomial \( f_2(x)=-1 + 2x^2 \) (where the fractional iterates become power series instead...).
And if I take any higher-indexed Chebychev-polynomial as the base-function , the iterate is simply computable by the cosh/acosh-mechanism
\( f_b^{\circ h}(x) = \cosh(b^h \operatorname{acosh}(x)) \)
(And for my initial question using sinh/asinh it is simply the same except with other polynomials)
Should have seen this before... -
Gottfried
[update]: And -oh wonder- this is just related to a question in MSE where I was involved this days without knowing that this two questions are in the same area; I just added the information about the cosh/arccosh-composition there... :-) http://math.stackexchange.com/questions/...965#490965
Of course the basic Chebychev-polynomials are "the-functions-of which-acosh(x)-is-the-Schröderfunction".
Let for instance \( f_2(x) = T_2(x) = -1 + 2x^2 \) then of course the iterates \( f_2^{\circ 2}(x) = T_2(T_2(x)) = 1 -8x^2+8x^4 \) and the iterates to any height h are \( f_2^{\circ h}(x)= T_2^{\circ h}(x) \) which can then be evaluated/interpolated using the cosh/acosh-pair as \( f_2^{\circ h}(x) = \cosh(2^h \operatorname{acosh}(x)) \) .
So my question is then -very simple- answered for "base=2" that the "more common" function "in our toolbox" is the quadratic polynomial \( f_2(x)=-1 + 2x^2 \) (where the fractional iterates become power series instead...).
And if I take any higher-indexed Chebychev-polynomial as the base-function , the iterate is simply computable by the cosh/acosh-mechanism
\( f_b^{\circ h}(x) = \cosh(b^h \operatorname{acosh}(x)) \)
(And for my initial question using sinh/asinh it is simply the same except with other polynomials)
Should have seen this before... -
Gottfried
[update]: And -oh wonder- this is just related to a question in MSE where I was involved this days without knowing that this two questions are in the same area; I just added the information about the cosh/arccosh-composition there... :-) http://math.stackexchange.com/questions/...965#490965
Gottfried Helms, Kassel