Continued fractions closed form - missed out members in July 1999
Bhushit Joshipura
back in July 1999.
Most of them were known to the community. However, I did not receive
comments on a couple of them.
Can anyone throw light now?
************
epsilon = 1+2/(3+4/(5+6/(7+...)))
=f(0) where f(n) = (2n+1)+[2(n+1)/f(n+1)]
************
phi = 0+1/(2+3/(4+5/(6+...)))
= f(0) where f(n) = 2n+[(2n+1)/f(n+1)]
************
Thanks in advance,
-Bhushit (not a mathematician)
se...@btinternet.com
> epsilon = 1+2/(3+4/(5+6/(7+...)))
> =f(0) where f(n) = (2n+1)+[2(n+1)/f(n+1)]
>
1/(e^(1/2)-1) ~ 1.541494...
G. A. Edgar
<se...@btinternet.com> wrote:
J. Rec. Math. 7 (1974) 152--153
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
Robert Israel
> I mailed questions about closed forms about many continued fractions
> back in July 1999.
> Most of them were known to the community. However, I did not receive
> comments on a couple of them.
> Can anyone throw light now?
>
> ************
> epsilon = 1+2/(3+4/(5+6/(7+...)))
> =f(0) where f(n) = (2n+1)+[2(n+1)/f(n+1)]
(already answered)
> ************
>
> phi = 0+1/(2+3/(4+5/(6+...)))
> = f(0) where f(n) = 2n+[(2n+1)/f(n+1)]
Write f(n) = p(n)/p(n+1). Then we get the recurrence
p(n) = 2 n p(n+1) + (2 n + 1) p(n+2)
which Maple can solve, giving the closed form
p(n) = -(-1)^n*(-GAMMA(n-1/2)*p(0)-I*GAMMA(n-1/2)*exp(-1/2)*Pi^(1/2)*2^(1/2)
+2*GAMMA(n-1/2)+GAMMA(n-1/2)*exp(-1/2)*Pi^(1/2)*erf(1/2*I*2^(1/2))*2^(1/2)*I
+2^(1/2)*Pi^(1/2)*exp(-1/2)*GAMMA(n-1/2,-1/2)*I)/GAMMA(n-1/2)
= (-1)^(n+1) (a - p(0) + b (Gamma(n-1/2, -1/2)/Gamma(n-1/2) - 1)
where a = 2 - exp(-1/2) sqrt(2 pi) erfi(1/sqrt(2))
and b = sqrt(2 pi) i exp(-1/2). Now Gamma(n-1/2, -1/2)/Gamma(n-1/2) - 1
goes to 0 rapidly as n -> infinity, so the only way to avoid f(n) -> -1
is p(0) = a, and then p(1) = exp(-1/2) sqrt(2 pi) erfi(1/sqrt(2)) and
f(0) = sqrt(2/pi) exp(1/2)/erfi(1/sqrt(2)) - 1. This is approximately
0.3797319547409956329.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
David W. Cantrell
> Bhushit Joshipura <josh...@gmail.com> writes:
>
> > I mailed questions about closed forms about many continued fractions
> > back in July 1999.
> > Most of them were known to the community. However, I did not receive
> > comments on a couple of them.
> > Can anyone throw light now?
> >
> > ************
> > epsilon = 1+2/(3+4/(5+6/(7+...)))
> > =f(0) where f(n) = (2n+1)+[2(n+1)/f(n+1)]
>
> (already answered)
>
> > ************
> >
> > phi = 0+1/(2+3/(4+5/(6+...)))
> > = f(0) where f(n) = 2n+[(2n+1)/f(n+1)]
>
> Write f(n) = p(n)/p(n+1). Then we get the recurrence
> p(n) = 2 n p(n+1) + (2 n + 1) p(n+2)
> which Maple can solve, giving the closed form
>
> p(n) =
> -(-1)^n*(-GAMMA(n-1/2)*p(0)-I*GAMMA(n-1/2)*exp(-1/2)*Pi^(1/2)*2^(1/2)
> +2*GAMMA(n-1/2)+GAMMA(n-1/2)*exp(-1/2)*Pi^(1/2)*erf(1/2*I*2^(1/2))*2^(1/2
> = (-1)^(n+1) (a - p(0) + b (Gamma(n-1/2, -1/2)/Gamma(n-1/2) - 1)
>
> where a = 2 - exp(-1/2) sqrt(2 pi) erfi(1/sqrt(2))
> and b = sqrt(2 pi) i exp(-1/2). Now Gamma(n-1/2, -1/2)/Gamma(n-1/2) - 1
> goes to 0 rapidly as n -> infinity, so the only way to avoid f(n) -> -1
> is p(0) = a, and then p(1) = exp(-1/2) sqrt(2 pi) erfi(1/sqrt(2)) and
> f(0) = sqrt(2/pi) exp(1/2)/erfi(1/sqrt(2)) - 1. This is approximately
> 0.3797319547409956329.
Excellent. I will submit that closed form, on your behalf, to be added at
<http://www.research.att.com/~njas/sequences/A113014>.
(Note: Neil is away now, and presumably will not be processing new comments
again until mid-Sept.)
David
tommy1729
that remainds me of my sequence about continued fractions together with robert.g.wilson
very similar.
tommy1729