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An interesting result occurs, if e<>1, thus if rescaling is allowed.
I tried to find one e, for which the resulting powerseries are linear scalings of each other, such that

g0(x)=c * g1(x)

This is possible.

Assume again, that, denoting the fixpoints p and q, such that
f(x) = x + (x - p) (x - q)
Denote also the substitution
z = x' = x/e - d
and the backsubstitution
x" = (x + d)*e
Recall, that for a fixpoint p, d and e must be selected that p = e*d

If I select
Code:
´     e0 = 1 + (p-q)    d0=p/e0          // first fixpoint     e1 = 1 + (q-p)    d1=q/e1          // second fixpoint
then the resulting G0 and G1-matrices provide the same powerseries with a linear scaling.

Example
Code:
´     using fixpoint p                     using fixpoint q                                   e0 = 1 + (p-q)                       e1 = 1 + (q-p)                                   z0 =  x/e0 - d0 = (x-p)/e0           z1 =  x/e1 - d1 = (x-q)/e1                                                                                                          g0(z) = e0 z^2 + (1+(p-q))z          g1(z) = e1 z^2 + (1+(q-p))z

The matrices and eigenmatrices for g0 and g1 are as follows:
Code:
´ W0                                       W1                 1        0       0    0     0     0  |  1            0         0      0      0       0   0        1       0    0     0     0  |  0            1         0      0      0       0   0        2       1    0     0     0  |  0           -2         1      0      0       0   0      8/3       4    1     0     0  |  0         24/5        -4      1      0       0   0     24/7    28/3    6     1     0  |  0     -1176/95      68/5     -6      1       0   0  416/105  368/21   20     8     1  |  0  204768/6175  -4176/95  132/5     -8       1   ...                                      ... D0                                       D1     1      1/2     1/4  1/8  1/16  1/32  |  1          3/2       9/4   27/8  81/16  243/32 W0^-1                                    W1^-1   1        0       0    0     0     0  |  1            0         0      0      0       0   0        1       0    0     0     0  |  0            1         0      0      0       0   0       -2       1    0     0     0  |  0            2         1      0      0       0   0     16/3      -4    1     0     0  |  0         16/5         4      1      0       0   0  -352/21    44/3   -6     1     0  |  0       416/95      52/5      6      1       0   0  2048/35  -384/7   28    -8     1  |  0   32768/6175   2048/95  108/5      8       1   ...                                      ... --------------------------------------- ------------------------------------------------- G0                                      G1   1        0       0    0     0     0  |  1            0         0      0      0       0   0      1/2       0    0     0     0  |  0          3/2         0      0      0       0   0      1/2     1/4    0     0     0  |  0          3/2       9/4      0      0       0   0        0     1/2  1/8     0     0  |  0            0       9/2   27/8      0       0   0        0     1/4  3/8  1/16     0  |  0            0       9/4   81/8  81/16       0   0        0       0  3/8   1/4  1/32  |  0            0         0   81/8   81/4  243/32   ...                                      ...

The additionally interesting thing is, that we get by this column-scaled binomial-matrices for G0 and G1
X = G0 * dV(2) = G1* dV(2/3)
Code:
X=   1  .  .  .  .   .  .  .   0  1  .  .  .   .  .  .   0  1  1  .  .   .  .  .   0  0  2  1  .   .  .  .   0  0  1  3  1   .  .  .   0  0  0  3  4   1  .  .   0  0  0  1  6   5  1  .   0  0  0  0  4  10  6  1
so maybe I have just introduced some triviality.

Anyway- this does not (yet?) provide a linear scaling for the half-powers. Perhaps one can find an x->x' substitution, which provides such a linear scaling - but this is only an idea, don't know, whether this is even possible.

Gottfried
Again I didn't see some simplification...

Here it is:
Code:
´  using      g(z)= e z^2 + (1+(p-q)) z    setting    e = (1+(p-q))
this is
Code:
´     g(z)= e z (z +  1)

second power of g(z) (not "iterate"!)
Code:
´   (g(z))^2 = e^2 z^2 (z+1)^2 = e^2 z^2 (c(2,0)z^2 + c(2,1)z + c(2,2) )

where c(a,b) is binomial(a,b) and this gives a binomial-matrix column-shifted accordingly to powers of g.

Can we describe the half-iterate from this?
Another problem were now, to find a functional relation between the selection of fixpoints, shifts - and possibly of the startparameter x- , and the result-differences, something like

f°h(x) = g0°h(x')" + v(p,h,x) = g1°h(x')" + v(q,h,x)

This looks a bit like the correction, that I've to introduce with my infinite alternating series of increasing powertowers, where

AU_b(x) = sum(h=0,inf,(-1)^h*expd_b° h(x))
AL_b(x) = sum(h=1,inf,(-1)^h*expd_b°-h(x))
where epxd_b(x) = b^x - 1 the decremented iterated exponential

and where I conjectured AU + AL = 0
but actually AU + AL - v(b,x) = 0

Perhaps there is a relation between these problems...
Gottfried Wrote:I tried to find one e, for which the resulting powerseries are linear scalings of each other, such that

g0(x)=c * g1(x)

This is possible.

Assume again, that, denoting the fixpoints p and q, such that
f(x) = x + (x - p) (x - q)
Denote also the substitution
z = x' = x/e - d
and the backsubstitution
x" = (x + d)*e
Recall, that for a fixpoint p, d and e must be selected that p = e*d

If I select
Code:
´     e0 = 1 + (p-q)    d0=p/e0          // first fixpoint     e1 = 1 + (q-p)    d1=q/e1          // second fixpoint
then the resulting G0 and G1-matrices provide the same powerseries with a linear scaling.

Ok, short verification:
\begin{align*}
g_0(x)&=f(x'')'\\
&=(x''+(x''-p)(x''-q))'\\
&=(xe_0+p + (xe_0+p-p)(xe_0+p-q))'\\
&=\frac{xe_0+e_0d + (xe_0)(xe_0+p-q)}{e_0} - d \\
&= x+x(xe_0+p-q)\\
&= x^2e_0 + x(1+p-q) = e_0(x^2+x)
\end{align*}

$g_1(x)=e_1(x^2+x)$
right. Thats indeed interesting.

Quote:Anyway- this does not (yet?) provide a linear scaling for the half-powers

Yes and thats the critical point here. There is no such law like
$(cf)^{\circ t}=c^tf^{\circ t}$
but at least we have traced back the regular iteration at different fixed points to the regular iteration of different scalings. Does this also work for higher order polynomials?
Ok, I've fiddled some days with this case of iteration. Now new information, but lots of paper with interesting graphs...

bo198214 Wrote:but at least we have traced back the regular iteration at different fixed points to the regular iteration of different scalings. Does this also work for higher order polynomials?

Hmm, I haven't done this, because even if we get the coefficients linearly scaled this does not give obvious and special helpful interesting information when iterated. Also I doubt, that it can be done with higher polynomials - it was just computing a +/- deviance d of same abolute value of a sort of mean-value for the two fixpoints. If we have three or more, like f(x) = x + (x-t0)(x-t1)(x-t2) then in general there are no common d's with same absolute value - so I think, this will be impossible.

However, this led me to rethink the formula derived from symbolic eigensystem-decomposition of the b^x-1 - iteration. Written as function of height parameter h it is something like $f_{b,x}(h) = a_{1_{x,u}}*u^h + a_{2_{x,u}}*u^{2h} + a_{3_{x,u}}*u^{3h} + ...$ (where u=log(b) and the a_k are functions of x and u, resp are constant for given x and u) which made me fiddle with the derivatives, sums of f(h) with increasing h (again) and so on. If I have news, I'll post it in the tetra-series thread.

Gottfried
Well, to not to lose all of the impules that this fiddling gave I'll add some more comments.

a) the example function. I considered the next iterate of f(x)=-1/16+x+x^2 f2(x) = x^4 + 2*x^3 + 15/8*x^2 + 7/8*x - 31/256
as base function; this can be rewritten in fixpoint notation
f2(x) = x + (x-1/4)(x+1/4)(x -1 + sqrt(15/16)*I)(x -1 - sqrt(15/16)*I)
So f2 has four fixpoints, two of them are complex. And to investigate the half-iterate of this would allow to shift by 1/4,1/2, -1+sqrt(15/16)*I, 31/16 and compare the repective solutions.
I didn't do this yet, but would be interested in a discussion of this anyway.

b) the decremented iterated exponential f(x) = b^x-1 , f°h(x) = f°(h-1)(f(x)).
From the reformulation as function of h, where b and x are assumed to be constant,

$f_{b,x}(h) = a_{1_{x,u}}*u^h + a_{2_{x,u}}*u^{2h} + a_{3_{x,u}}*u^{3h} + ...$
(where u=log(b) and the a_k are functions of x and u, resp are constant for given x and u)
gives some ideas concerning moduli and cyclotomic functions; however since the coefficients a are non-integer (surely irrational) I don't see yet serious applications for modulus-considerations.
However, for cyclotomic h (abs(h)=1, arg(h)/Pi= rational) this looks, as if the powertowers $f_{b,x}(exp(k/n 2\pi i))$ of cyclotomic heights are periodic with k. Also series of such powertowers should be interesting.

Also I'm trying to see, whether this description gives new insights for the inconsistency of the matrix-approach and the serial summation for the alternating tetra-series (of increasing negative heights), but I don't see this so far.

Gottfried
I should also add a remark, which I've made already. This is the link to the Euler-paper E190 which seem to deal with a false series for logarithm, in the sense:
s_a = 1
s_a°(h+1) = a*s_a°h
for integer h - which means, this series mimics the logarithm of a^h - for integer h exclusively.
This series looks much like the series, which we get if we use the fixpoint-shift and evaluate iterates: exact results with integer heights, false results with algebraic heights. Perhaps its series-inverse is even a closer related series to our tetration-series (as given in the symbolic eigensystem-decomposition), and we are on a false track, when employing the fixpoint-shift.

Gottfried
Gottfried Wrote:then the resulting G0 and G1-matrices provide the same powerseries with a linear scaling.

I just see that this also works for exponentials $b^x$, if there are two fixed points $a_1$ and $a_2$ then we have
$b^{x+a_i}-a_i=b^{a_i} b^x- a_i = a_i b^x - a_i = a_i(b^x-1)$.
So the development at both fixed points are linear scalings of each other and both scalings of $b^x-1$.
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