Re: [math-fun] computing non-rigorously defined quantities
At 08:41 AM 7/2/2013, James Propp wrote:
For some thoughts about the enterprise of computing quantities one hasn't rigorously defined, see
http://mathoverflow.net/questions/135536/procedure-based-as-opposed-to-defin... (which I hope will elicit interesting responses over the next few days).
Date: Tue, 02 Jul 2013 14:16:11 -0600 From: rcs@xmission.com
Some of the funsters' flow work fits here:
Hah! "Fits," but comfortably?
Another, even more surprising example, is f(x) = sqrt2^x. This has fixed points at x = 2 & 4, and is well-behaved (nearly linear, smooth, etc.) near the fixed points and in the interval between.
sqrt(2) is close to 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. expm1 is not a true exponential, but the further up the flow you *start*, the closer it fits the "towers" of e. (log1p(x) = ln(1 + x), inverse of expm1.) + -inf ^ + 0.61301989 ^ | exp ln | | expm1 log1p | v -0.00000000 ^ v 0.84599770 ^ | exp ln | | expm1 log1p | v 1.00000000 ^ v 1.33030161 ^ | exp ln | | expm1 log1p | v 2.71828183 ^ v 2.78218395 ^ | exp ln | | expm1 log1p | v 15.15426224 ^ v 15.15426250 ^ | exp ln | | expm1 log1p | v 3814279.10476017 +----v 3814279.10476017 + | | +------------------------------------------+ The outer and inner U's, including turning around short of infinity, are procedures to map between, not only the two sequences, but between any point on the expm1^t flow and the implied exp^t flow. The difference between the two columns at x is like e^-x, so you get like e^^n bits of precision in n steps, which you can carry down the other side (er up, in the diagram). There isn't a simple mapping between flows of exponentials of different bases, but this technique makes flows for e and e^(1/e), and I think with some tweaks, for any base (including yet another answer for sqrt(2)!). So, I start with a function with a single fixed point at zero and extract something from that, but then go almost out to infinity and back a different way to cook it. And the mapping must make a complete hash of complex numbers. And even if there's a way to fix that... Personally I'm not happy with just these procedures or numbers or curves, although they're something I've been able to do and they're pretty. I wish they were somehow close or related to something with a real reason to live. I wonder about Beauty in math, the smell that's supposed to tell you you're on to something. --Steve + -inf ^ + 0.61301989 ^ exp | | ln expm1 | | log1p v -0.00000000 ^ v 0.84599770 ^ exp | | ln expm1 | | log1p v 1.00000000 ^ v 1.33030161 ^ exp | | ln expm1 | | log1p v 2.71828183 ^ v 2.78218395 ^ exp | | ln expm1 | | log1p v 15.15426224 ^ v 15.15426250 ^ exp | | ln expm1 | | log1p v 3814279.10476017 +------------------v 3814279.10476017 + | | +--------------------------------------------------------+
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.
participants (2)
-
Dan Asimov -
Steve Witham