On Dec 29, 12:31 pm, William Stein <wst...@gmail.com> wrote:> On Dec
28, 2011 9:05 PM, "shreevatsa" <shreevatsa.pub...@gmail.com> wrote:> >
On Dec 29, 3:29 am, John Cremona <john.crem...@gmail.com> wrote:> > >
This follows very easily from Stirling's Formula. But I suppose the>
> > original question was not so much "what is the answer" as "can I
get> > > the answer from system [xyz] without having to think"!> > >
Yes!> > > After all, anything Sage can do a human can probably do with
enough> > patience and time. The point of the software is to make it
easy.> > (Not meant as flaimbait) I can think of a lot of
counterexamples to this,> at least in the context of number theory
problems that involve counting,> which would be impossible to do in a
lifetime because humans are just way> too slow...
Yes, I agree. I thought of this actually... so I waffled a bit with
"enough patience and time", despite realizing the time required can be
more than a lifetime. :-) My fault.Either way, I think my point stands
that even if we can do something by hand, it would still be useful for
Sage to have the ability to do it.What needs to be done here for this
particular case, does anyone know?
To answer my original question (I just thought of this), this is how I
think one can get the answer in Sage with minimum human effort.
1. Look up Stirling's series to as many terms as desired. E.g.
n! = √(2πn) (n/e)^n (1 + 1/(12n) + 1/(288n^2) - 139/(51840n^3) - 571/
(2488320n^4) + O(1/n^5))
So this is √(2πn) (n/e)^n multiplied by a power series in 1/n. And
Sage can handle power series.
2. In Sage:
sage: var('n') n sage: f = sqrt(2*pi*n) * (n/e)^n sage:
R.<x> = PowerSeriesRing(QQ) sage: g = 1 + 1/12 * x + 1/288 * x^2 -
139/51840 * x^3 - 571/2488320 * x^4 + O(x^5)
We've written the power series separately, so that factorial(n) = f(n)
* g(1/n). To check:
sage: (f(1000)*g(1/1000)).n() 4.02387260077094e2567 sage:
factorial(1000).n() 4.02387260077094e2567
Cool. Now, binomial(n, n/2) = factorial(n) / factorial(n/2)^2 = f(n)/
f(n/2)^2 * g(1/n)/g(2/n)^2. So to get a series expansion for
binomial(n, n/2)/2^n:
sage: simplify(f(n)/f(n/2)^2/2^n) sqrt(2)/(sqrt(pi)*sqrt(n))
and
sage: g(x)/g(2*x)^2 1 - 1/4*x + 1/32*x^2 + 5/128*x^3 -
21/2048*x^4 + O(x^5)
This shows that binomial(n, n/2)/2^n is
√(2/(πn)) (1 - 1/(4n) + 1/(32n^2) + 5/(128n^3) - 21/(2048n^4) + O(1/
n^5))
which coincides with the answer given by WA. Much more convenient than
multiplying and dividing power series by hand.(If the Stirling series
were stored in Sage, it would avoid even the first step of typing it
in.)
Thanks again for Sage,Shreevatsa
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