Road testing Ansus' continuum product formula

[mike3]

Yes, but only three iterations. You've already seen the graphics. The second iteration I simply found symbolically and the third I found numerically.

So you've never tried a numeric procedure up to lots and lots of iterations, then? I think, however, that I may have figured out what the problem is.

First off, we start with the formula:
\(
\log_a \frac{f'_a(x)}{f'_a(0)(\ln a)^{x}} =\sum_{n=1}^{\infty} \frac{f^{(n-1)} (0)}{n!} (B_n(x)-B_n(0))
\)

We can write

\( B_n(x) = \sum_{k=0}^{n} {n \choose k} B_{n - k} x^k \),

and

\( B_n(x) - B_n(0) = \sum_{k=1}^{n} {n \choose k} B_{n - k} x^k \).

So the rhs of the formula becomes

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

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

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

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. 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.

I'd guess that to make it work one would need to employ some sort of divergent summation method to extend the range of applicability of the formula (i.e. to sum up \( b_k = \sum_{n=1}^{\infty} \frac{f^{(n-1)}(0)}{n!} {n \choose k} B_{n - k} \) when the coefficients \( \frac{f^{(n-1)}(0)}{n!} \) decay slowly), or else find an alternate method of computing solutions to the continuum-sum equation (any ideas? How could Mma's NDSolve have done it???).