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 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). --Dan On 2013-07-19, at 10:21 PM, Steve Witham wrote:
B = e^(1/e) = ~1.444667 whose exponential has one ("double?") fixed point, B^e = e and, you know, B^x is an honest exponential. If you scale and translate B^x a bit, (B^(ex + e) - e)/e = e^x - 1 = expm1(x) = x + x^2/2! + x^3/3! + ... expm1(0) = 0
If you set the initial x to (say) 1/2, then the flow, expm1^t(1/2) has a Taylor series (of t) with nice rational coefficients and an interval around zero where it converges. Except I'm getting the coefficients from sympy and don't see a pattern beyond knowing what I asked it.