Section 7.2.1.4: Generating All Partitions

Exercise 45. [HM21] [HM21] Compute the asymptotic value of p(n−1)/p(n), with relative error O(n−2).

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Solution

Let $p(n)$ denote the partition function. From the Hardy–Ramanujan asymptotic formula (first Rademacher term suffices at the required precision),

$$ p(n)=\frac{1}{4n\sqrt{3}}\exp!\left(a\sqrt{n}\right)\left(1+O!\left(e^{-c\sqrt{n}}\right)\right), \qquad a=\pi\sqrt{\frac{2}{3}}, $$

so the error is exponentially small and does not affect any expansion in powers of $n^{-1/2}$.

Hence

$$ \frac{p(n-1)}{p(n)}

\frac{n}{n-1}\exp!\bigl(a(\sqrt{n-1}-\sqrt{n})\bigr)\left(1+O!\left(e^{-c\sqrt{n}}\right)\right). $$

Set

$$ \sqrt{n-1}=\sqrt{n}\left(1-\frac{1}{n}\right)^{1/2}. $$

Using the binomial expansion,

$$ \left(1-\frac{1}{n}\right)^{1/2}

1-\frac{1}{2n}-\frac{1}{8n^{2}}-\frac{1}{16n^{3}}+O(n^{-4}), $$

so

$$ \sqrt{n-1}

\sqrt{n} -\frac{1}{2\sqrt{n}} -\frac{1}{8n^{3/2}} -\frac{1}{16n^{5/2}} +O(n^{-7/2}). $$

Therefore

$$ \Delta:=\sqrt{n-1}-\sqrt{n}

-\frac{1}{2\sqrt{n}} -\frac{1}{8n^{3/2}} -\frac{1}{16n^{5/2}} +O(n^{-7/2}). $$

Define

$$ t_1=-\frac{a}{2\sqrt{n}},\quad t_2=-\frac{a}{8n^{3/2}},\quad t_3=-\frac{a}{16n^{5/2}}. $$

Then

$$ a\Delta=t_1+t_2+t_3+O(n^{-7/2}). $$

Expansion of the exponential

First,

$$ \exp(t_1)

1-\frac{a}{2\sqrt{n}}+\frac{a^2}{8n}-\frac{a^3}{48n^{3/2}}+\frac{a^4}{384n^2}+O(n^{-5/2}). $$

Next,

$$ \exp(t_2)=1-\frac{a}{8n^{3/2}}+O(n^{-3}), $$

and $t_3$ contributes only beyond $n^{-5/2}$ in the final product.

Multiplying,

$$ \exp(a\Delta)

1-\frac{a}{2\sqrt{n}}+\frac{a^2}{8n} -\left(\frac{a^3}{48}+\frac{a}{8}\right)n^{-3/2} +\left(\frac{a^4}{384}+\frac{a^2}{16}\right)n^{-2} +O(n^{-5/2}). $$

Prefactor expansion

$$ \frac{n}{n-1}=\frac{1}{1-1/n}=1+\frac{1}{n}+\frac{1}{n^2}+O(n^{-3}). $$

Final multiplication

Let

$$ S= 1+s_1 n^{-1/2}+s_2 n^{-1}+s_3 n^{-3/2}+s_4 n^{-2}, $$

where

$$ s_1=-\frac{a}{2},\quad s_2=\frac{a^2}{8},\quad s_3=-\left(\frac{a^3}{48}+\frac{a}{8}\right),\quad s_4=\frac{a^4}{384}+\frac{a^2}{16}. $$

Then

$$ \frac{p(n-1)}{p(n)}

\left(1+\frac{1}{n}+\frac{1}{n^2}\right)S+O(n^{-5/2}). $$

Collecting terms up to $n^{-2}$ gives

$$ \frac{p(n-1)}{p(n)}

1 +s_1 n^{-1/2} +(s_2+1)n^{-1} +(s_3+2s_1)n^{-3/2} +(s_4+s_2+1)n^{-2} +O(n^{-5/2}). $$

Substituting $a=\pi\sqrt{\frac{2}{3}}$ yields the explicit expansion

$$ \frac{p(n-1)}{p(n)}

1 -\frac{\pi}{\sqrt{6n}} +\left(\frac{\pi^2}{12}+1\right)\frac{1}{n} -\left(\frac{\pi^3}{36\sqrt{6}}+\frac{\pi}{\sqrt{6}}\right)\frac{1}{n^{3/2}} +\left(\frac{\pi^4}{288}+\frac{\pi^2}{24}+1\right)\frac{1}{n^2} +O(n^{-5/2}). $$

Thus the asymptotic value of $p(n-1)/p(n)$, with relative error $O(n^{-2})$, is

$$ \boxed{ 1 -\frac{\pi}{\sqrt{6n}} +\left(\frac{\pi^2}{12}+1\right)\frac{1}{n} -\left(\frac{\pi^3}{36\sqrt{6}}+\frac{\pi}{\sqrt{6}}\right)\frac{1}{n^{3/2}} +\left(\frac{\pi^4}{288}+\frac{\pi^2}{24}+1\right)\frac{1}{n^2} }. $$

This completes the solution. ∎