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

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, 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, 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

Offset: 1

Author

Clark Kimberling, Oct 03 2011

Keywords

• nonn
• cons

Comments

Let L be the least x>0 satisfying cos(x)=1/sqrt(1+x^2).

Then cos(x) < 1/sqrt(1+x^2) for 0

Consequently,

cos(x) < 1/sqrt(1+x^2) < (1/x)sin(x) for 0

(see A196504). Equivalently,

cos(1/x) < x/sqrt(1+x^2) < x sin(1/x) for x>0.49291...

(see A196505).

Examples

			L=4.9131804394348836888378206685945355668...
1/L=0.203534149076564439697574222397395290289996941...
		

Links

• Index entries for transcendental numbers.

Crossrefs

Cf. A196500, A196503, A196504.

Programs

Mathematica

Plot[{Cos[x], 1/Sqrt[1 + x^2]}, {x, 0, 8}]
t = x /.FindRoot[1/Sqrt[1 + x^2] == Cos[x], {x, 4, 5}, WorkingPrecision -> 100]
RealDigits[t]   (* A196502 *)
1/t
RealDigits[1/t] (* A196503 *)

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