Natural logarithm of 2 [HP-15C/HP-42S/Free42 & others]
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11-08-2019, 03:18 PM
(This post was last modified: 11-09-2019 02:29 AM by Albert Chan.)
Post: #15
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RE: Natural logarithm of 2 [HP-15C/HP-42S/Free42 & others]
Using Pi and AGM to get log(2): see http://rnc7.loria.fr/brent_invited.pdf, page 32
\(\large log(2^n) = \frac{\pi/2}{AGM(1, \frac{4}{2^n})} \left(1 + O(\frac{1}{2^{2n}}) \right)\) Big enough n so that error term may be ignored. Assume O(1/4^n) = 1/4^n : 1/4^n < 1e-12 n ≥ 12 / log10(4) ≥ 19.93 Casio FX-115MS, with n = 20: A=1 B=2^-18 C=A+B : B=√AB : A=C/2 : A-B ==== → 0.4980487824 ==== → 0.2197274566 ==== → 0.0525527223 ==== → 0.0030466068 ==== → 0.0000102395 ==== → 0.0000000001 Convergence close to quadratic. With final A,B, we have log(2) ≈ pi/(n(A+B)) = 0.693147180558, under-estimated 2e-12 Update: More precise calculations shows that n=20 over-estimated log(2) by 2e-12 Below suggested relative error term O(1/4^n) ≈ 1/4^(n-1) python> from gmpy2 import * python> f = lambda n: const_pi() / (2 * n * agm(1, 4/2**n)) python> for n in range(20, 25): print n, "%.15g" % f(n) ... 20 0.693147180562285 21 0.693147180560532 22 0.693147180560093 23 0.693147180559982 24 0.693147180559955 |
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