OFFSET
0,1
COMMENTS
dilog(phi-1) = polylog(2, 2-phi) = sum((2-phi)^k/k^2 , k =1 ..infinity) = sum((1 - 2*sin(Pi/10))^(2*k)/k^2, k=1..infinity) = Pi^2/15 - (log(phi-1))^2 = Pi^2/15 - (2/5)*log(phi-1)*(log(2-phi) + log(phi-1)/2).
See the Jolley reference pp. 66-69, (360)(e), and the Abramowitz-Stegun link, p. 1004, eqs. 27.7.3 - 27.7.6 with x = phi-1, solving for dilog(x) = f(x).
REFERENCES
L. B. W. Jolley, Summation of Series, Dover, 1961.
LINKS
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
FORMULA
dilog(phi-1) = polylog(2, 2-phi) = Sum_{k>=1} (2-phi)^k/k^2 = Sum_{k>=1} (1 - 2*sin(Pi/10))^k/k^2.
EXAMPLE
0.42640880616209618209...
MAPLE
phi := (1+sqrt(5))/2 ; dilog(phi-1) ; evalf(%) ; # R. J. Mathar, Jun 10 2024
MATHEMATICA
RealDigits[PolyLog[2, 2 - GoldenRatio], 10, 120][[1]] (* Amiram Eldar, May 27 2023 *)
PROG
(PARI) polylog(2, 2 - (1+sqrt(5))/2) \\ Gheorghe Coserea, Sep 30 2018
(PARI) sumpos(k=1, (1 - 2*sin(Pi/10))^k/k^2) \\ Gheorghe Coserea, Sep 30 2018
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Wolfdieter Lang, Jun 16 2014
STATUS
approved