Exact and Unique solution for base e^(1/e)

[jaydfox]

But I'm only looking at the first 15 or so terms of the series. It seems pretty well behaved and very well-defined, but maybe I'm missing something?

What exactly did you try?
Did you experiment with the iterations of \( e^x-1 \)? *wondering*

Indeed I did. I found that if you take the first several iterations, and look strictly at the a_n's for a given n, they obey a very well-defined (n-1)th-order polynomial:

\( \begin{eqnarray}
f(x) & = & e^x-1 \\
f^{\circ 0}(x) & = & 0\left(\frac{1}{0!}\right)\ +\ 1\left(\frac{x}{1!}\right)\ +\ 0\left(\frac{x^2}{2!}\right)\ +\ 0\left(\frac{x^3}{3!}\right)\ +\ 0\left(\frac{x^4}{4!}\right)\ +\ \dots \\
f^{\circ 1}(x) & = & 0\left(\frac{1}{0!}\right)\ +\ 1\left(\frac{x}{1!}\right)\ +\ 1\left(\frac{x^2}{2!}\right)\ +\ 1\left(\frac{x^3}{3!}\right)\ +\ 1\left(\frac{x^4}{4!}\right)\ +\ \dots \\
f^{\circ 2}(x) & = & 0\left(\frac{1}{0!}\right)\ +\ 1\left(\frac{x}{1!}\right)\ +\ 2\left(\frac{x^2}{2!}\right)\ +\ 5\left(\frac{x^3}{3!}\right)\ +\ 15\left(\frac{x^4}{4!}\right)\ +\ \dots \\
\end{eqnarray}
\)

It's easier to make out the patterns when the coefficients are laid out in a matrix, without all the other stuff to confuse things:

\( \begin{array}{c|ccc}
& 0 && 1 && 2 && 3 && 4 && 5 && 6 && 7 & \dots\\
\hline
0 & 0 && 1 && 0 && 0 && 0 && 0 && 0 && 0 & \dots\\
1 & 0 && 1 && 1 && 1 && 1 && 1 && 1 && 1 & \dots\\
2 & 0 && 1 && 2 && 5 && 15 && 52 && 203 && 877 & \dots\\
3 & 0 && 1 && 3 && 12 && 60 && 358 && 2471 && 19302 & \dots\\
4 & 0 && 1 && 4 && 22 && 154 && 1304 && 12915 && 146115 & \dots\\
5 & 0 && 1 && 5 && 35 && 315 && 3455 && 44590 && 660665 & \dots\\
6 & 0 && 1 && 6 && 51 && 561 && 7556 && 120196 && 2201856 & \dots\\
\vdots & \vdots && \vdots && \vdots && \vdots && \vdots && \vdots && \vdots && \vdots & \ddots
\end{array} \)

If you try to find an equation for the mth iteration, looking only at the terms in the mth row, you'll be a sad camper. But if you look at any particular column n, you should easily be able to verify that it's terms are part of a sequence defined by an (n-1)th order polynomial.

Armed with this knowledge, it should be trivial to calculate the first 15-20 constants for the iteration m=0.5, for example. With a sufficiently powerful math library, one should be able to derive any desired number of terms in the sequence.

On a sidenote, I'm using Microsoft Excel with the XNumbers library, but Mathematica or Maple, etc., would be preferable. Alas, I'm not a student (self-study apparently doesn't count), and I can't afford the non-student version.

Anyway, the polynomials themselves should behave very well. As such, the m=0.5 iteration should converge much faster than the m=3 iteration, for example, at least in the vicinity of 0. The radius of convergence should be greater than 0, and regardless of how small it might be, any radius of convergence greater than 0 guarantees that analytic extension is possible.

However, I've only looked at the first dozen or so polynomials (for the first dozen or so terms of the sequences). There seems to be a well-defined formula for determining the coefficient of the highest degree term of the [/i]n[/i]th polynomial, so it should be possible to prove convergence. I haven't "proven" convergence per se, but I can see the foundation for a proof in the making.