"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):
> 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.
> "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):
> > 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.
>"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):
>> 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
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.
Rob Johnson <r...@trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font
> 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):
> >> 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
> 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. -- Robert Israel isr...@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
> > 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):
> > >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
> > 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.
Now it looks to me like Psi(m+t) - Psi(m+1-t) has only odd powers of m in its asymptotic series. Indeed Psi(m+t) - Psi(m+1-t) = 2 sum_{j=0}^infty Psi(2j+1, m+1/2)/(2j+1)! (t-1/2)^(2j+1)
where Psi(k,x) is the k'th derivative of Psi(x).
Now asymptotically Psi(m) = ln(m) - 1/(2m) + (terms in even powers of m) from the Euler-Maclaurin formula and the fact that bernoulli(k) = 0 when k > 1 is odd. Using the identity Psi(m+1/2) = 2 Psi(2 m) - Psi(m) - 2 ln(2) the terms in 1/m cancel, so Psi(m+1/2) = ln(m) + (terms in even powers of m). Taking derivatives, we find that Psi(k,m+1/2) has only odd powers of m when k is an odd integer and only even powers of m when k is an even integer. Up to some questions of interchanging asymptotics and sums, this completes the proof. -- Robert Israel isr...@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
