7
$\begingroup$

I am needing to use the asymptotic formula for the partition number, $p(n)$ (see here for details about partitions).

The asymptotic formula always seems to be written as,

$ p(n) \sim \frac{1}{4n\sqrt{3}}e^{\pi \sqrt{\frac{2n}{3}}}, $

however I need to know the order of the omitted terms, (i.e. I need whatever the little-o of this expression is). Does anybody know what this is, and a reference for it? I haven't been able to find it online, and don't have access to a copy of Andrews 'Theory of Integer Partitions'.

Thank you.

$\endgroup$
  • $\begingroup$ I believe the asymptotic nature of the Hardy-Ramanujan formula, notwithstanding its use to get exact values of $p(n)$, means that a "little-o" notation for omitted terms would be misplaced. $\endgroup$ – hardmath Jul 5 '11 at 16:14
  • $\begingroup$ @Hardmath, I'm about to answer my own question (!), but also justify that there is a little-o representation, since in fact if f is asymptotically equivalent to g, then f = (1 + o(1))g... So as pointed out, I've now answered my original question... silly me. $\endgroup$ – owen88 Jul 5 '11 at 17:18
  • 3
    $\begingroup$ Okay, I thought perhaps the exact convergent series given by Rademacher (1937) that refines the Hardy-Ramanujan formula (which forms the first term of the series) and its order of convergence might be of interest. G. Andrews has a chapter about this in his book Theory of Integer Partitions. $\endgroup$ – hardmath Jul 5 '11 at 17:34
10
$\begingroup$

The original paper addresses this issue on p. 83:

$$ p(n)=\frac{1}{2\pi\sqrt2}\frac{d}{dn}\left(\frac{e^{C\lambda_n}}{\lambda_n}\right) + \frac{(-1)^n}{2\pi}\frac{d}{dn}\left(\frac{e^{C\lambda_n/2}}{\lambda_n}\right) + O\left(e^{(C/3+\varepsilon)\sqrt n}\right) $$ with $$ C=\frac{2\pi}{\sqrt6},\ \lambda_n=\sqrt{n-1/24},\ \varepsilon>0. $$

If I compute correctly, this gives $$ e^{\pi\sqrt{\frac{2n}{3}}} \left( \frac{1}{4n\sqrt3} -\frac{72+\pi^2}{288\pi n\sqrt{2n}} +\frac{432+\pi^2}{27648n^2\sqrt3} +O\left(\frac{1}{n^2\sqrt n}\right) \right) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.