From: Dan Asimov <dasimov@earthlink.net>

What method or formula are you using to get the Taylor coefficients of expm1^t at x = 1/2, where expm1(x) := e^x - 1 ??? (I don't know what sympy is or what query you have submitted to it.)

I've sent a more detailed answer to Dan. The short story is, I started with N (around 18) terms of the Taylor series of f(x) = e^((x+e)/e) - e P = x + x^2/(2! e) + x^3/(3! e^2) ... I generated M polynomials, for 0 through M-1 iterations of P, each up to N terms. (I'm forgetting whether M needs to be N+1 or N-1 right now.) Visualize an array of the coefficients, with the iterations as rows. Say t is the row number, starting at zero. The column for the constant term is all zero. the coefficient of x is all 1, constant down the column x^2, a linear progression x^3, a quadratic x^4, a cubic ... I inferred polynomials of t from the columns, giving a polynomial in both x and t. I substituted in x = .5, reducing the whole thing to a polynomial in t. The general way to calculate the function is to use the polynomial for t mod 1, then apply either f or its inverse repeatedly as needed.

I ask particularly because I have read that e^x - 1 (and hence e^(x/e)) does not have any non-integer analytic iterates. See I. N. Baker, Zusammensetzungen ganzer Functionen, Math. Zeitschr. Bd. 69, pp. 121-163 (1958).

I haven't read the article, haven't even tried complex values of t. For reals less than about 13, it seems well-behaved. --Steve