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)