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A335822 Given the two curves y = (1 + exp(x))/2 and y = (1 + exp(x))/(1 + exp(x/2)), draw a line tangent to both. This sequence is the decimal expansion of the y-coordinate of the point at which the line touches y = (1 + exp(x))/2. 4
8, 5, 9, 2, 4, 4, 4, 7, 6, 4, 2, 1, 3, 9, 9, 3, 1, 3, 2, 9, 3, 7, 3, 0, 4, 7, 1, 2, 9, 6, 2, 3, 2, 3, 1, 0, 8, 1, 7, 0, 1, 2, 3, 3, 4, 1, 7, 8, 9, 9, 5, 7, 0, 3, 1, 7, 2, 6, 4, 2, 7, 8, 8, 0, 8, 6, 7, 6, 6, 5, 6, 2, 6, 7, 4, 4, 8, 0, 8, 7, 8, 2, 4, 2, 9, 7, 8, 6, 0, 1, 7, 4, 9, 1, 1, 0, 1 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET
0,1
LINKS
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.
Eric Weisstein's World of Mathematics, Shapiro's cyclic sum constant.
FORMULA
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)).
This sequence gives the decimal expansion of B = (1 + exp(b))/2 = (1 + exp(A335809))/2.
mu = A086278 = (1 + exp(b)*(1 - b))/2 = B - (B - 1/2)*log(2*B - 1) and A335810 = c.
The constant B - 1/2 = exp(b)/2 = 0.35924447642... gives the slope of the common tangent.
EXAMPLE
0.8592444764213993132937304712...
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)))/2
CROSSREFS
Cf. A086277 (constant lambda), A086278 (constant mu), A243261 (Gauchman's constant), A245330 (2*lambda), A335809 (-b), A335810 (c), A335825 (y-coordinate for c).
Sequence in context: A074071 A100126 A330111 * A233033 A244810 A273985
KEYWORD
nonn,cons
AUTHOR
Petros Hadjicostas, Jun 25 2020
STATUS
approved

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Last modified September 12 21:40 EDT 2024. Contains 375855 sequences. (Running on oeis4.)