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%I #16 Sep 27 2022 15:34:22
%S 3,7,9,2,7,3,4,9,6,4,9,7,3,8,8,0,7,2,6,7,2,2,1,5,3,4,4,5,2,2,4,4,6,4,
%T 3,2,0,6,9,2,1,3,1,8,2,8,2,0,2,6,5,4,9,8,3,3,4,4,9,4,1,0,6,8,9,1,2,7,
%U 4,0,6,8,5,5,0,4,7,8,6,8,8,1,6,0,3,1,6,5,8,7,0,0,7,6,7,7,8,8,6
%N Decimal expansion of Pi/(2 + 2*Pi).
%C The number Pi/(2 + 2*Pi) is the least x > 0 such that sin(x) = cos(Pi*x).
%C If b and c are distinct real numbers, the solutions of sin(bx) = cos(cx) are x = (k - 1/2)*Pi/(b + c), where k runs through the integers. Thus, if b > 0 and c > 0, the least solution x > 0 is Pi/(2*b + 2*c), so that this is also the least x > 0 for which sin(c*x) = cos(b*x). Related sequences, each with a Mathematica program which includes a graph:
%C ...
%C b.....c.......sequence........x
%C 1.....2.......A019673........ x = Pi/6
%C 1.....3.......A019678........ x = Pi/8
%C 1.....4.......(A000796)/10... x = Pi/10
%C 1.....Pi......A197682........ x = Pi/(2+2*Pi)
%C 1.....2*Pi....A197683........ x = Pi/(2+4*Pi)
%C 1.....1/Pi....A197684........ x = Pi^2/(2+2*Pi)
%C 1.....2/Pi....A197685........ x = Pi^2/(4+2*Pi)
%C 1.....Pi/2....A197686........ x = Pi/(2+Pi)
%C 1.....Pi/3....A197687........ x = 3*Pi/(6+2*Pi)
%C 1.....Pi/4....A197688........ x = 2*Pi/(4+Pi)
%C 1.....Pi/6....A197689........ x = 3*Pi/(6+Pi)
%C 2.....3.......(A000796)/10... x = Pi/10
%C 2.....Pi......A197690........ x = Pi/(4+2*Pi)
%C 2.....2*Pi....A197691........ x = Pi/(4+4*Pi)
%C 2.....1/Pi....A197692........ x = Pi^2/(2+4*Pi)
%C 2.....2/Pi....A197693........ x = Pi^2/(4+4*Pi)
%C 2.....Pi/2....A197694........ x = Pi/(4+Pi)
%C 3.....Pi......A197695........ x = Pi/(2+2*Pi)
%C 3.....2*Pi....A197696........ x = Pi/(6+4*Pi)
%C 3.....1/Pi....A197697........ x = Pi^2/(2+6*Pi)
%C 3.....2/Pi....A197698........ x = Pi^2/(4+6*Pi)
%C 3.....Pi/2....A197699........ x = Pi/(6+Pi)
%C 1/2...Pi......A197700........ x = Pi/(1+2*Pi)
%C 1/2...2*Pi....A197701........ x = Pi/(1+4*Pi)
%C 1/2...1/Pi....A197724........ x = Pi^2/(2+Pi)
%C 1/2...2/Pi....A197725........ x = Pi^2/(4+Pi)
%C 1/2...Pi/2....A197726........ x = Pi/(1+Pi)
%C 1/2...Pi/4....A197727........ x = 2*Pi/(2+Pi)
%C 1/3...Pi/3....A197728........ x = 3*Pi/(2+2*Pi)
%C 1/3...Pi/6....A197729........ x = 3*Pi/(2+Pi)
%C 2/3...Pi/6....A197730........ x = 3*Pi/(4+Pi)
%C 1/4...Pi......A197731........ x = 2*Pi/(1+4*Pi)
%C 1/4...Pi/2....A197732........ x = 2*Pi/(1+2*Pi)
%C 1/4...Pi/4....A197733........ x = 2*Pi/(1+Pi)
%C 1/5...Pi/5....10*A197691..... x = 5*Pi/(2+2*Pi)
%C 1/6...Pi/6....A197735........ x = 3*Pi/(1+Pi)
%C 1/8...Pi/8....A197736........ x = 4*Pi/(1+Pi)
%H <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>
%e 0.37927349649738807267221534452244643...
%t b = 1; c = Pi;
%t t = x /. FindRoot[Sin[b*x] == Cos[c*x], {x, .3, .4}]
%t N[Pi/(2*b + 2*c), 110]
%t RealDigits[%] (* A197682 *)
%t Simplify[Pi/(2*b + 2*c)]
%t Plot[{Sin[b*x], Cos[c*x]}, {x, 0, Pi}]
%o (PARI) 1/(2/Pi+2) \\ _Charles R Greathouse IV_, Sep 27 2022
%Y Cf. A197683.
%K nonn,cons
%O 0,1
%A _Clark Kimberling_, Oct 17 2011
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