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A319568 Given the two curves y = exp(-x) and y = 2/(exp(x) + exp(x/2)), draw a line tangent to both. This sequence is the decimal expansion of the (negated) x-coordinate of the point at which the line touches y = 2/(exp(x) + exp(x/2)). 6

%I #53 Jun 23 2020 09:00:01

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

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

%U 7,9,0,8,2,7,9,5,1,1,6,3,9,3,8,9,8,6,0,5,6,8,2,0,0,2,8,5,0,3,6,3,5,1,8,9,6,7

%N Given the two curves y = exp(-x) and y = 2/(exp(x) + exp(x/2)), draw a line tangent to both. This sequence is the decimal expansion of the (negated) x-coordinate of the point at which the line touches y = 2/(exp(x) + exp(x/2)).

%C From _Petros Hadjicostas_, Jun 01 2020: (Start)

%C The calculations in this sequence and in A319569 are needed for the estimation of the Shapiro cyclic sum constant lambda = A086277 = phi(0)/2 = A245330/2. This was done in Drinfel'd (1971).

%C Here phi(x) is the convex hull of y = exp(-x) and y = 2/(exp(x) + exp(x/2)); i.e., phi(x) = 2/(exp(x) + exp(x/2)) for x <= b; phi(x) = exp(-c) + ((2/(exp(b) + exp(b/2)) - exp(-c))/(b - c)) * (x - c) for b <= x <= c; and phi(x) = exp(-x) for x >= c. (For b <= x <= c, we have the equation of the line segment tangent to both curves.)

%C As stated below, b = -0.200811... (current sequence) and c = 0.15519... (A319569).

%C It follows that phi(0) = exp(-c) - c * ((2/(exp(b) + exp(b/2)) - exp(-c))/(b - c)) (where the y-axis crosses the line segment). Or by using the tangent line at x = c to the curve y = exp(-x), we find phi(0) = exp(-c)*(c + 1). Or by using the tangent line at x = b to the curve y = 2/(exp(x) + exp(x/2)), we may get a third formula for phi(0).

%C By using the above information, we get a system of two equations with two unknowns (b and c). See the PARI program below that may be used to calculate b.

%C Similar calculations were done by Elbert (1973) for the Shapiro cyclic sum constant mu = psi(0) = A086278. The corresponding curves are y = (1 + exp(x))/2 and y = (1 + exp(x))/(1 + exp(x/2)), while the corresponding x-coordinates at the tangent points are b' = -0.33060... and c' = 0.38875... (not in the OEIS yet). Here psi(x) is the convex hull of these two curves (and it becomes a line segment tangent to both curves for b' <= x <= c').

%C _Eric W. Weisstein_, in the link below, has a summary of the above discussion (with contributions by _Steven Finch_). (End)

%D Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 3.1, Shapiro-Drinfeld Constant, p. 209.

%H V. G. Drinfel'd, <a href="https://doi.org/10.1007/BF01316982">A cyclic inequality</a>, Mathematical Notes of the Academy of Sciences of the USSR, 9 (1971), 68-71.

%H Á. Elbert, <a href="https://doi.org/10.1007/BF02276104">On a cyclic inequality</a>, Periodica Mathematica Hungarica, 4 (1973), 163-168.

%H Á. Elbert, <a href="https://akjournals.com/view/journals/10998/4/2-3/article-p163.xml">On a cyclic inequality</a>, Periodica Mathematica Hungarica, 4 (1973), 163-168.

%H Petros Hadjicostas, <a href="/A086277/a086277.pdf">Plot of the curves y = exp(-x) and y = 2/(exp(x) + exp(x/2)) and their common tangent</a>, 2020.

%H R. A. Rankin, <a href="https://doi.org/10.2307/3608356">2743. An inequality</a>, Mathematical Gazette, 42(339) (1958), 39-40.

%H H. S. Shapiro, <a href="https://www.jstor.org/stable/2307617">Proposed problem for solution 4603</a>, American Mathematical Monthly, 61(8) (1954), 571.

%H H. S. Shapiro, <a href="https://www.jstor.org/stable/2306671">Solution to Problem 4603: An invalid inequality</a>, American Mathematical Monthly, 63(3) (1956), 191-192; counterexample provided by M. J. Lighthill.

%H B. A. Troesch, <a href="https://doi.org/10.1090/S0025-5718-1989-0983563-0">The validity of Shapiro's cyclic inequality</a>, Mathematics of Computation, 53 (1989), 657-664.

%H Eric Weisstein's MathWorld, <a href="http://mathworld.wolfram.com/ShapirosCyclicSumConstant.html">Shapiro's Cyclic Sum Constant</a>.

%F b = -0.2008110658618... .

%F c = 0.1551949747226... (A319569).

%F exp(c) = (1 + exp(b/2))^2/(2 + exp(-b/2)).

%e -0.2008110658618...

%o (PARI) c(b) = b + exp(b/2)/(2*exp(b)+exp(b/2))

%o solve(b=-2,2, exp(-c(b))*(1-b+c(b))-2/(exp(b)+exp(b/2))) \\ _Petros Hadjicostas_, Jun 02 2020

%Y Cf. A086277, A086278, A245330, A319569.

%K nonn,cons

%O 0,1

%A _Seiichi Manyama_, Sep 23 2018

%E More terms from _Vaclav Kotesovec_, Sep 26 2018

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