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Decimal expansion of log((1+e)/2).
0

%I #16 Aug 16 2020 11:33:47

%S 6,2,0,1,1,4,5,0,6,9,5,8,2,7,7,5,2,4,6,3,1,7,6,3,3,7,3,5,0,9,6,7,9,0,

%T 7,3,8,3,9,7,7,9,9,5,1,3,1,0,0,9,3,1,2,0,5,9,8,3,8,3,5,0,5,3,4,3,8,0,

%U 1,2,7,9,5,3,9,4,9,2,5,0,5,3,3,0,3,3,7,0,7,9,2,8,0,8,7,3,7,0,3,8

%N Decimal expansion of log((1+e)/2).

%C Positive matrices do not always behave like positive numbers: for instance 0 <= A <= B does not imply e^A <= e^B [e^A stands here for matrix exponential of A, and 0 <= A <= B means that A and B-A are positive matrices, a positive matrix being a matrix which has nonnegative entries]. The counterexample given by Paul Halmos is A = e^{{1/2, 1/2}, {1/2, 1/2}} = {{(1+e)/2, (e-1)/2}, {(e-1)/2, (1+e)/2}} and B = {{e^x, 0}, {0, e^y}}, with x < log((1+e)/2).

%D Paul Halmos, "Problems for Mathematicians, Young and Old", Dolciani Mathematical Expositions, 1991, Solution to problem 9Q (Exponential inequality) p. 273.

%F Equals Integral_{x=0..1} 1/(exp(-x) + 1) dx. - _Amiram Eldar_, Aug 16 2020

%e 0.620114506958277524631763373509679073839779951310093120598383505343801279539...

%t RealDigits[Log[(1 + E)/2], 10, 100][[1]]

%K nonn,cons

%O 0,1

%A _Jean-François Alcover_, Jul 22 2013