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A196502 Decimal expansion of the least positive x satisfying cos(x)=1/sqrt(1+x^2).

%I A196502 #8 Feb 11 2025 13:32:35
%S A196502 4,9,1,3,1,8,0,4,3,9,4,3,4,8,8,3,6,8,8,8,3,7,8,2,0,6,6,8,5,9,4,5,3,5,
%T A196502 5,6,6,8,4,7,6,1,1,8,4,5,0,6,6,1,5,5,0,5,6,1,4,2,1,8,5,2,0,5,5,1,2,7,
%U A196502 3,3,9,7,8,7,6,1,0,9,6,1,7,0,3,6,7,9,9,7,6,6,3,4,6,8,3,8,2,6,9,8
%N A196502 Decimal expansion of the least positive x satisfying cos(x)=1/sqrt(1+x^2).
%C A196502 Let L be the least x>0 satisfying cos(x)=1/sqrt(1+x^2).
%C A196502 Then cos(x) < 1/sqrt(1+x^2) for 0<x<L=4.91318...
%C A196502 Consequently,
%C A196502 cos(x) < 1/sqrt(1+x^2) < (1/x)sin(x) for 0<x<2.02876...
%C A196502 (see A196504).  Equivalently,
%C A196502 cos(1/x) < x/sqrt(1+x^2) < x sin(1/x) for x>0.49291...
%C A196502 (see A196505).
%H A196502 <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>.
%e A196502 L=4.9131804394348836888378206685945355668...
%e A196502 1/L=0.203534149076564439697574222397395290289996941...
%t A196502 Plot[{Cos[x], 1/Sqrt[1 + x^2]}, {x, 0, 8}]
%t A196502 t = x /.FindRoot[1/Sqrt[1 + x^2] == Cos[x], {x, 4, 5}, WorkingPrecision -> 100]
%t A196502 RealDigits[t]   (* A196502 *)
%t A196502 1/t
%t A196502 RealDigits[1/t] (* A196503 *)
%Y A196502 Cf. A196500, A196503, A196504.
%K A196502 nonn,cons
%O A196502 1,1
%A A196502 _Clark Kimberling_, Oct 03 2011