At K(k), the quarter period, the even derivatives of cn(x) vanish by symmetry. The special thing that happens at k = 1/sqrt(2) is that the third derivative vanishes. Thus the graph of cn(x) differs from a straight line by terms of 5-th order. -- Gene
________________________________ From: Bill Gosper <billgosper@gmail.com> To: math-fun@mailman.xmission.com Sent: Friday, December 23, 2011 12:12 PM Subject: [math-fun] DLMF fungous inflection
(This was too easy for the kids, but it feels good when you find the words an Authority missed.)
See http://dlmf.nist.gov/22.3.F2.mag and http://dlmf.nist.gov/22.3.F3.mag "For cn(x,k) the curve for k=1/sqrt2=0.70710… is a boundary between the curves that have an inflection point in the interval 0≤x≤2K(k), and its translates, and those that do not;..."
a) why don't they just say "at x=K(k)" vs "in the interval 0≤x≤2K(k)"?
b) Huh? cn(x,k) always has an inflection point there! But *something* happens at k=1/sqrt2. What should they have said? --rwg Caution: DLMF uses K(modulus) vs K(parameter), so if you plot this with Macsyma or Mma, the magic k = 1/2, not 1/sqrt2. In either case, K=4 (1/4)!^2/Sqrt[Pi] ~ 1.85407 (This K property is imperfectly analogous to the graphical definition of Halphen's constant, about 69% into http://gosper.org/thetpak.html .) _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun