Road testing Ansus' continuum product formula

[bo198214]

We can reverse the summation signs to get this in the form of a power series in x^k:

\(
\sum_{n=1}^{\infty} \sum_{k=1}^{n} \frac{f^{(n-1)}(0)}{n!} {n \choose k} B_{n - k} x^k = \sum_{k=1}^{\infty} \left(\sum_{n=1}^{\infty} \frac{f^{(n-1)}(0)}{n!} {n \choose k} B_{n - k}\right) x^k
\)

Yes, I did a similar derivation here and Ansus confirmed the formula here. Can you check whether our derivations coincide? (I did not consider the constant term though.)

So each coefficient of the power series for the continuum sum is itself a series, which may or may not converge. It turns out that for some functions, the coefficients for the continuum sum do not converge even if there the other series has nonzero radius of convergence.

Good that you raise this topic. I already asked Ansus about it, but it seems he is not so interested in the convergence stuff, or simply has no answers.
It also seems that Markus Mueller defines his fractional sum in a different way. Maybe this is due to exactly those convergence issues.

I suspect, that for any function that is not entire, i.e. whose power series has finite radius of convergence, and this includes tetration (note the presence of a singularity at z = -2), this will not work. If you try the above sum formula with the coefficients for log(1 + x), you'll see it won't work, despite the fact that the continuum sum for log(1 + x) exists and has a power series expansion at 0.

Would be interesting to have some theorem on when the coefficients of the extended/continuum sum converge. Can you prove your claim about the sum of entire functions?