> > In article <20080706103816.687...@newsreader.com>,
> > David W. Cantrell <DWCantr...@sigmaxi.net> wrote:
> > >"sa...@kt-algorithms.com" <sa...@kt-algorithms.com> wrote:
> > >> An interesting type of approximation to the central binomial
> > >> coefficient is of the form:

> > >>   C(2n, n) ~=  4^n / sqrt( Rm(n) )

> > >> Rm being a rational function of max. degree m+1 (m>0):

> > >>   Rm(n) = (pi n^(m+1) + sum(i=0,m) Pi n^i) / (n^m + sum(i=0,m-1) Qi
> > >>   n^i)

> > >> and where the 2m+1 coefficients are defined by simply requiring exact
> > >> results for 2m>=n>=0; note that C(0,0)=1 implies that Q0=P0.

> > >Hello, Knud!

> > >Yes, your form of approximation is nice.

> > >FWIW, here are simple upper and lower bounds for C(2n, n).

> > >Upper bound: 4^n / sqrt(pi(n + 1/4))                             (Ub)

> > >Lower bound: Ub * (1 - 1/(8(n + 1/4))^2)                         (Lb)

> > >Your approximations are precise for small n, while the relative errors
> > >of
> > >my bounds at n = 0 are roughly 13% and -15%, resp. But for large n,
> > >those
> > >bounds make reasonably good approximations. For example, at n = 50,
> > >their
> > >relative errors are about 6*10^-6 and -4*10^-10, resp.

> > Using Stirling's assymptotic series, we get that the central binomial
> > coefficient is assymptotically

> >        4^n            1       1         5          21
> >     ---------- ( 1 - --- + ------- + -------- - --------- - ... )
> >     sqrt(pi n)       8 n   128 n^2   1024 n^3   32768 n^4

> > which, using the first 3 terms, gives

> > n   C(2n,n) approximation
> > -   ------- -------------
> > 1   2       1.99229446690301
> > 2   6       5.99660115228404
> > 3   20      19.9965102093602
> > 4   70      69.9947702088935
> > 5   252     251.990291239197

> > Multiply by sqrt(1 + 1/(4n)) / sqrt(1 + 1/(4n)) to get

> >           4^n               1         1         3
> >     --------------- ( 1 - ------ + ------- - -------- - ... )
> >     sqrt(pi(n+1/4))       64 n^2   128 n^3   8192 n^4

> > Substituting m = n+1/4, we get

> >         4^m             1         21
> >     ----------- ( 1 - ------ + -------- - ... )
> >     sqrt(2pi m)       64 m^2   8192 m^4

> > Using one and two terms in the series, we get your approximations.
> > When n = 0, the third term throws things awry as is often the case
> > with assymptotic expansions.

> > The interesting part is that out to the 1/m^50 term, the series is
> > even.  This would indicate that C(2m-1/2,m-1/4) 4^{-m} sqrt(m) is
> > even.  I have not had a chance to look into this yet.

> That can't be literally true.  For example, C(2m-1/2,m-1/4) 4^(-m) sqrt(m)
> has poles at m = 1/4 - k/2 for positive integers k.