From: Raymond Manzoni Subject: Re: Continued Fraction [0z;1z,2z,3z,...] Date: Wed, 25 Aug 1999 01:29:34 +0200 Newsgroups: sci.math Keywords: Lehmer's -- continued fractions with quotients in arithmetic prog Hi, You are right. There's a general formula in Abramowitz "Handbook of mathematical functions" Dover (page 363) which claims that : [0; 2s/z, 2(s+1)/z, 2(s+2)/z, ...] = I_s(z) / I_{s-1}(z) This book is in S. Finch references. Perhaps he found this continued fraction there... Hope it helped, Raymond ============================================================================== From: "David Petry" Subject: Re: Continued Fraction [0z;1z,2z,3z,...] Date: Tue, 24 Aug 1999 20:09:22 -0700 Newsgroups: sci.math r.e.s. wrote >"Numerical exploration" supports the conjecture that >more generally: > >[0z;1z,2z,3z,...] = I1(2/z)/I0(2/z) for all complex z, The proof is actually rather easy. Start with the differential equation y = sy' + xy'' (differentiation with respect to x) Then repeated differentiation gives y^{k} = (s+k)y^{k+1} + xy^{k+2} Note that y/y' = s + x/(y'/y'') = s + x/(s+1 + x/(y''/y''')) etc. leading to a continued fraction if you choose the proper solution to the differential equation. From there, just a little manipulation of the continued fraction, and finding the explicit solution of the differential equation, gives you the result you're after. ============================================================================== From: "G. A. Edgar" Subject: Re: Continued Fraction [0z;1z,2z,3z,...] Date: Wed, 25 Aug 1999 07:58:41 -0400 Newsgroups: sci.math In article <7pv82l\$e84\$1@nntp1.atl.mindspring.net>, r.e.s. wrote: > At http://www.mathsoft.com/asolve/constant/cntfrc/cntfrc.html > a work by Lehmer is referenced for this interesting fact: > > [0;1,2,3,...] = I1(2)/I0(2) > > where [a0;a1,a2,a3,...] = a0+1/(a1+1/(a2+1/(a3+... > and > I1(z) = sum( (z/2)^2k+1 /(k!(k+1)!), k=0,1,2,...), > I0(z) = sum( (z/2)^2k /(k! k! ), k=0,1,2,...), > are modified Bessel functions, Ik(z)=J(iz)/i^k. > > "Numerical exploration" supports the conjecture that > more generally: > > [0z;1z,2z,3z,...] = I1(2/z)/I0(2/z) for all complex z, > > in the sense that the right- left-hand sides are equal > for all non-zero complex z, and both are undefined at > z=0, sharing there a limit equal to 1. > Yes, denominators in arithmetic progression is covered by Lehmer [Scripta Math. 29 (1973) 17--24]. For any b and any nonzero a, the continued fraction [b; b+a, b+2a, b+3a,...] has value I[b/a - 1](2/a) ------------------- I[b/a](2/a) where the Bessel function I[t](z) is infinity ----- (t + 2 m) \ (z/2) ) --------------------------------- / GAMMA(m + 1) GAMMA(t + m + 1) ----- m = 0 The proof is based on the recursive formula for the convergents of the continued fraction and the known recurrence for the Bessel functions. Substitute b=0 to obtain your conjecture; note I[-1](z) = I[1](z) . > Also, letting f(z)=[0z;1z,2z,3z,...], it appears that > f(z*)=f*(z) wrt to complex conjugation. (A very rough > picture is at http://rs.1.home.mindspring.com/cfz.jpg) The formula f(z*)=(f(z))* is true for any analytic function that is real for real z. > I don't have access to the Lehmer reference, and > am just wondering if this is discussed there, or how > one might go about proving the conjecture. -- Gerald A. Edgar edgar@math.ohio-state.edu Department of Mathematics telephone: 614-292-0395 (Office) The Ohio State University 614-292-4975 (Math. Dept.) Columbus, OH 43210 614-292-1479 (Dept. Fax)