andydude Wrote:Well, this is approximately what I was thinking. If \( f(x) = b^{b^x} \), then we are trying to solve \( f(x_0) = x_0 \) and \( f(x_1) = x_1 \) where \( x_0 \ne x_1 \)right? OK, so I started with a series expansion, and this is easier if you deal with a single exponential rather than an iterated exponential, so I found the series expansions:Hi Andrew -
I actually do not understand the job, that x0 and x1 shall do. But it gives a mental association to me, which may be helpful for your considerations, too.
From the eigensystem-analysis of decremented iterated exponentiation
\( U^{(h)}_t(x), \) where \( U^{(1)}_t(x) = t^x -1, \)
and \( u=\log(t) \)) I get a bivariate polynomial in u and u^h, where h indicates the "height" (see below).
Since on the other hand, I found, that the iterated exponentiation
\( T^{(h)}_b(x), \) (with x=1 means our usual tetration)
can be converted into
\( U^{(h)}_t(x), \)
by scaling and shifting of x we may use these polynomials also for T-tetration.
We have, using b=t^(1/t), (t complex for b>eta) the substitution x'=x/t-1, and inverse substitution x"=(x+1)*t,
\( T^{(h)}_b(x) = (U^{(h)}_t(x')) '' \)
so we may apply the polynomials to x' and use the result y' by applying the inverse substitution y"
Since we have denominators of the powerseries for U() , which generate singularities for all complex u, where abs(u)=1 (which means for instance, b=e^(e^-1) = eta,u=1, or b=(e^-1)^e = beta , u=-1) these cases come clearly out as special cases. However, by the construction of the polynomials, for integer heights h the denominators can be cancelled by the same factors in the numerators, and we have a powerseries with 0/0-terms, or, when cancelled, surely an equivalent solution to the logarithm-approach (but I didn't check this so far).
Here is the powerseries for \( U^{(h)}_t(x), \) (sorry , when editing the formula I overlooked one occurence of f(), which should be rewritten as U())
which I assume is the "ultimate" description for the tetration-problem (assuming continuation of tetration via powerseries-expansion) from where other considerations about the behaviour of tetration can be derived (at least for real u, which means for bases b 0<b<eta)
Perhaps your x0,x1-idea may be inserted into this polynomial and be resolved this way. If x0=t (the fixpoint of b), then x' = x/x0-1 in the above formula, and the terms of the powerseries for U(x) expand and mix then with binomial weights. (don't know, whether this agrees with your idea, however).
I can provide the matrix of polynomially coefficients at the powers of x, where one might then insert values for u and for u^h up to size 64x64. Unfortunately the polynomials for each term of the powerseries grow binomially c(r,2) with their rowindex r, so the size of this matrix is huge (200 MB in text-format): for 96 terms of the powerseries (which was my intended goal) one even had a polynomial of degree ~4400 in the 96'term and if I count the number of all occuring coefficients at powers of u individually I get about 10 million coefficients for that number of x^k-polynomials... urrgs...
Gottfried
Gottfried Helms, Kassel