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A086278 Decimal expansion of Shapiro's cyclic sum constant mu. 12
9, 7, 8, 0, 1, 2, 4, 7, 8, 1, 8, 6, 6, 4, 6, 2, 2, 0, 2, 0, 1, 8, 2, 7, 9, 5, 9, 9, 7, 8, 6, 8, 2, 6, 8, 0, 9, 3, 2, 5, 3, 8, 6, 3, 5, 3, 4, 5, 9, 1, 4, 1, 8, 0, 9, 4, 9, 5, 3, 0, 4, 2, 0, 8, 3, 4, 5, 9, 9, 4, 4, 9, 2, 5, 8, 0, 7, 1, 0, 6, 9, 7, 5, 0, 0, 5, 5, 6, 6, 8, 9, 8, 5, 2, 0, 3, 9, 2, 6, 5, 9, 2, 4 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET
0,1
COMMENTS
From Petros Hadjicostas, Jun 01 2020: (Start)
Based on the references, it seems that this constant was first defined by Elbert (1973). We have psi(0) = mu. Two auxiliary constants, b = -0.33060494... = -A335809 and c = 0.38755227... = A335810, are needed for the estimation of mu.
Here psi(x) is the convex hull of y = (1 + exp(x))/2 and y = (1 + exp(x))/(1 + exp(x/2)); i.e., psi(x) = (1 + exp(x))/2 for x <= b; psi(x) = (1 + exp(b))/2 + (((1 + exp(c))/(1 + exp(c/2)) - (1 + exp(b))/2)/(c - b)) * (x - b) for b <= x <= c; and psi(x) = (1 + exp(x))/(1 + exp(x/2)) for x >= c. (For b <= x <= c, we have the equation of the line segment tangent to both curves.)
It follows that mu = psi(0) = (1 + exp(b))/2 - b * (((1 + exp(c))/(1 + exp(c/2)) - (1 + exp(b))/2)/(c - b)) (where the y-axis crosses the line segment). Or by using the tangent line at x = b to the curve y = (1 + exp(x))/2, we find mu = psi(0) = (1 + exp(b))/2 - b * exp(b)/2. Or by using the tangent line at x = c to the curve y = (1 + exp(x))/(1 + exp(x/2)), we may get a third formula for mu = psi(0) in terms of c only.
Similar calculations were done by Drinfel'd (1971) for the Shapiro cyclic sum constant lambda = phi(0)/2 = A086277 = A245330/2. In this case, the corresponding curves are y = exp(-x) and y = 2/(exp(x) + exp(x/2)), while the corresponding x-coordinates at the tangent points are -A319568 = -0.20081... and A319569 = 0.15519... Here phi(x) is the convex hull of these two curves (and it becomes a line segment tangent to both curves for -A319568 <= x <= A319569).
Eric W. Weisstein, in the link below, has a summary of the above discussion (with contributions by Steven Finch). (End)
REFERENCES
Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 3.1, Shapiro-Drinfeld constant, p. 209.
LINKS
V. G. Drinfel'd, A cyclic inequality, Mathematical Notes of the Academy of Sciences of the USSR, 9 (1971), 68-71.
Á. Elbert, On a cyclic inequality, Periodica Mathematica Hungarica, 4 (1973), 163-168.
Á. Elbert, On a cyclic inequality, Periodica Mathematica Hungarica, 4 (1973), 163-168.
R. A. Rankin, 2743. An inequality, Mathematical Gazette, 42(339) (1958), 39-40.
H. S. Shapiro, Proposed problem for solution 4603, American Mathematical Monthly, 61(8) (1954), 571.
H. S. Shapiro, Solution to Problem 4603: An invalid inequality, American Mathematical Monthly, 63(3) (1956), 191-192; counterexample provided by M. J. Lighthill.
B. A. Troesch, The validity of Shapiro's cyclic inequality, Mathematics of Computation, 53 (1989), 657-664.
Eric Weisstein's World of Mathematics, Shapiro's cyclic sum constant.
FORMULA
From Petros Hadjicostas, Jun 23 2020: (Start)
Solve the following system of equations to find the x-coordinates of the two points where the common tangent touches the two curves:
exp(b) = (-exp(c/2) + 2*exp(c) + exp(3*c/2))/(1 + exp(c/2))^2 and
(exp(b)*(c - b + 1) + 1)*(1 + exp(c/2)) = 2*(1 + exp(c)).
Then the constant equals (1 + exp(b)*(1 - b))/2. (End)
It can be proved that this constant equals 1 plus the Gauchman constant (which is the negation of A243261); i.e., without negations, A086278 = 1 - A243261. - Petros Hadjicostas, Jul 04 2020
EXAMPLE
0.97801247818664622020182795997868268... = 1 - 0.02198752181335377979817204...
MATHEMATICA
eq = E^u + 2*E^(u + v/2) + E^(v/2) + E^(u + v) == 2*E^v + E^(3*v/2) && 2 + 2*E^(u + v/2) == 2*y + 2*E^v + E^u*(v - 2) && E^u*(u - v + 1) + 2*E^(u + v/2) + 1 == 2*E^v; mu = y /. FindRoot[eq , {{y, 1}, {u, -1/3}, {v, 1/3}}, WorkingPrecision -> 105]; RealDigits[mu, 10, 103] // First
PROG
(PARI)
default("realprecision", 200)
b(c) = log((-exp(c/2) + 2*exp(c) + exp(3*c/2))/(1 + exp(c/2))^2);
a = solve(c=-1, 1, (exp(b(c))*(c - b(c) + 1) + 1)*(1 + exp(c/2)) - 2*(1 + exp(c)));
(1 + exp(b(a))*(1 - b(a)))/2 \\ Petros Hadjicostas, Jun 23 2020
CROSSREFS
Sequence in context: A203079 A232128 A336081 * A081855 A019887 A163931
KEYWORD
nonn,cons
AUTHOR
Eric W. Weisstein, Jul 14 2003
EXTENSIONS
More terms from Jean-François Alcover, Jun 02 2014
STATUS
approved

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Last modified April 28 05:00 EDT 2024. Contains 372020 sequences. (Running on oeis4.)