A nice series for b^^h , base sqrt(2), by diagonalization

[bo198214]

Gottfried, we discussed that already in an much earlier post.
Your whole infinite matrix computation is just r e g u l a r i t e r a t i o n !!!

Let me make it clear again:

If we have the Bellmatrix B for b^x , then we cannot invert B for the infinite case. But

B = fS2F * P~ // both triagular, invertible

First Bell matrix behave antiisomorphic: let B[f] be the Bell matrix of f at 0, then
B[fog]=B[g]*B[f]

Now let us assign the variables:

B is the Bellmatrix of \( \exp \) at 0.
P~ is the Bellmatrix of \( \tau_1(x)=x+1 \)
fS2F is the Bellmatrix of \( \text{dexp}(x)=\exp(x)-1 \)
(for my convenience I omit the base sub script at exp and dexp)

As dexp has a fixed point at 0 fS2F is upper triangular. Your equation just corresponds to
\( \exp = \tau_1 \circ \text{dexp}(x) \)
\( \exp(x) = 1 + \exp(x) - 1 \)

But P (as well as then PInv) are the binomial-matrices and they perform addition when operating on the powerseries:

V(x)~ *PInv~ = V(x-1)~

exactly: \( (\tau_1)^{-1} = \tau_{-1} \), the inverse of \( x+1 \) is \( x-1 \).

Then rearranging the invertible P as PInv to the left

V(e^x)*PInv~ = V(x)~ * fS2F

then the invertible fS2F into fS1F to the left

V(e^x)*PInv~ * fS1F = V(x)~

where the product to construct B^-1 is forbidden *in the infinite case* but, applying the binomial-theorem by PInv

V(e^x - 1)~ * fS1F = V(x) ~

which is perfectly ok, and this leads then, since fS1F performs log(1+x)

Thats clear without touching any matrix:
V(e^x - 1) is the Bell matrix of dexp, and fS1F is the inverse of the Bell matrix of dexp:
\( \text{dexp}^{-1}\circ \text{dexp} = \text{id} \)

I do the same with the eigensystem-decomposition / Schröder-function. I found the fixpoint-shift with my matrix-notation by simply proceeding from the initial equation

V(x)~ * B = V(y)~

decomposing B into matrix-factors P, P^-1 and a triangular C

V(x)~ * P^-t ~ * C * P^t ~ = V(y)~

applying binomial-theorem with the -t'th power of P~ on rhs und lhs

V(x-t)~ * C = V(y - t)~ // implements shift by t = fixpoint

where then C is triangular and allows eigendecomposition providing exact values

\( \exp=\tau_{t}\circ \underbrace{\tau_t^{-1} \circ \exp \circ \tau_t}_{\text{texp}} \circ \tau_t^{-1} \)

C is the Bell matrix of texp.
\( \text{texp}(x)=\tau_t^{-1}\circ \exp\circ \tau_t (x) = \exp(x+t)-t = \exp(x)t -t = t(\exp(x)-1) \)
we see that texp has 0 as fixed point and hence is the Bell matrix triangular.
And the power can be taken exactly.

All in all- I use the "regular iteration with fixpoint-shift", but only as far as I can represent it coherently in terms of infinite matrices/known closed-form expressions for sums of infinite powerseries, which result by the implicte dot-products.

But I still dont see what it brings to rewrite an established method with matrices. Whether you compute it with matrices, with direct powerseries formulas, or with limit formulas (without powerseries), whether with Abel or super function, the result is always the same, its always regular iteration.

Thus I have the difficulties with b>eta, since the occuring complex-valued matrices C give unsatisfying powerseries, and I have not yet the remedy to deal with that series appropriately.

I gave the answer already in the previous post. The resulting power series \( t+\sum_{n=0}^\infty p_n (x-t)^n \) (where \( p_n \) are the coefficients obtained from \( C^h \)) has convergence radius |t| in your notation. It converges at most for \( x\in (0,2 \text{Re}(t)) \) on the real axis.

Yes, thanks! It's progressing diabetes, and sometimes I'm fitter, sometimes struck down, and in general just less powerful and regenerable than up to recent years. Just life..

For me it seems you should do something healthy not always brood about the same things. Spring is coming, go out, have a look at the blue sky, or at the young girls passing