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Given the partition function $P(n)$ and let $q_k=e^{-k\pi/5}$. What is the reason why,

$$\sum_{n=0}^\infty P(n) q_2^{n+1}\approx\frac{1}{\sqrt{5}}\tag{1}$$

$$\sum_{n=0}^\infty P(n) q_4^{n+1}\approx\Big(1-\frac{1}{5^{1/4}}\Big)^2\Big(\frac{1+\sqrt{5}}{4}\Big)\tag{2}$$

where the difference is a mere $10^{-14}$ and $10^{-29}$, respectively? (The form of $(2)$ seems to be more than coincidence.)

P.S. I forgot where I found $(2)$. Does anybody recall the paper where this approximation first appears?

Tito Piezas III
  • 60,745

3 Answers3

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We may observe that your sum is $$\sum_{n\geq 0} P(n) q^{n+1}=\frac{q}{f(-q)}\tag{1}$$ where $f(-q) =(q;q) _{\infty} $ is one of Ramanujan theta functions. And further it is known that $$\frac{f(-q)} {qf(-q^{25})}=\frac{1}{R(q^5)}-1-R(q^5)\tag{2}$$ is algebraic for $q=q_4=e^{-4\pi/5}$. Here $R(q) $ is the Rogers-Ramanujan continued fraction.

With $q^{25}=e^{-20\pi}\approx 5.1579\times 10^{-28}$ I think we can safely ignore the factor $f(-q^{25})$ as almost $1$. Your sum is then almost equal to reciprocal of $$\frac{1}{R(e^{-4\pi})}-1-R(e^{-4\pi})=e^{4\pi/5}\frac{f(-e^{-4\pi/5})}{f(-e^{-20\pi})}$$ The above constant is well known and evaluated in my blog post as $$\frac{\sqrt[4]{5}+1}{\sqrt[4]{5}-1}\cdot \sqrt{5}$$ and its reciprocal is $$\frac{1}{a^2}\cdot\frac{a-1}{a+1}$$ with $a=5^{1/4}$. It can be seen that this equals your expression $$\left(1-\frac{1}{\sqrt[4]{5}}\right)^2\frac{1+\sqrt{5}}{4}$$ We have $$\left(1-\frac{1}{a}\right)^2\cdot\frac{1+a^2}{4}=\frac{(a-1)^2}{a^2}\frac{1+a^2}{a^4-1}=\frac{a-1}{a^2(a+1)}$$

Paramanand Singh
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I have not derived the approximation (2) yet, but the approximation (1) is easy:

Set $q(\tau)=\mathrm{e}^{2\pi\mathrm{i}\tau}$. We will use the Dedekind eta function $$\eta(\tau) = q\left(\frac{\tau}{24}\right) \prod_{n=1}^\infty\left(1-q(n\tau)\right) = \frac{q\left(\frac{\tau}{24}\right)}{\sum_{n=0}^\infty P(n) q(n\tau)}$$

We also know the following modular forms property of $\eta(\tau)$: $$\eta(\tau) = \sqrt{\frac{\mathrm{i}}{\tau}}\eta\left(\frac{-1}{\tau}\right)$$

Set $\tau=\frac{\mathrm{i}}{5}$. Then $\frac{-1}{\tau} = 5\mathrm{i} = 25\tau$. Thus $$\begin{align} q(\tau) &= \mathrm{e}^{-2\pi/5} \\ q\left(\frac{-1}{\tau}\right) &= q(25\tau) = \mathrm{e}^{-10\pi} \approx 2\cdot10^{-14} \\ \eta\left(\frac{-1}{\tau}\right) &= \eta(25\tau) \approx q\left(\frac{25\tau}{24}\right) \\ \eta(\tau) &= \sqrt{5}\,\eta(25\tau) \approx \sqrt{5}\,q\left(\frac{25\tau}{24}\right) \end{align}$$ That is, the $\prod_{n=1}^\infty\cdots$ used in $\eta(25\tau)$ well approximates $1$.

Thus $$\begin{align} \sum_{n=0}^\infty P(n) q\left((n+1)\tau\right) &= \frac{q(\tau)\,q\left(\frac{\tau}{24}\right)}{\eta(\tau)} = \frac{q\left(\frac{25\tau}{24}\right)}{\eta(\tau)} \\ &\approx \frac{q\left(\frac{25\tau}{24}\right)} {\sqrt{5}\,q\left(\frac{25\tau}{24}\right)} = \frac{1}{\sqrt{5}} \end{align}$$ Approximation (2) seems to require a little more.

ccorn
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  • Thanks, ccorn! I just noticed that, let $q=e^{-2\pi/7}$, then, $$\sum_{n=0}^\infty P(n) q^{n+\color{red}{2}}\approx\frac{1}{\sqrt{7}}$$ which differs only by $10^{-15}$. Will the same analysis work? What other $q=e^{-2\pi/p}$ will do? – Tito Piezas III Nov 28 '13 at 03:08
  • For $p$ coprime to $6$, $m=\frac{p^2-1}{24}$ is an integer. Then for $\tau=\frac{\mathrm{i}}{p}$ we have $\frac{-1}{\tau}=p^2\tau$ and approximating $\sum_{n=0}^\infty P(n),q^{n+m}$ as shown will cancel the $q\left(\frac{p^2\tau}{24}\right)$. – ccorn Nov 28 '13 at 03:17
  • Ah, I see. I can't find an analogous approximation to $(2)$ for $q=e^{-k\pi/7}$ with $k \ne 2$ though. – Tito Piezas III Nov 28 '13 at 03:27
  • Yes, $(2)$ must be more involved, as the presence of $5^{1/4}$ shows. This cannot come from modular inversion, so there must be at least one more functional equation for $\eta$ involved. – ccorn Nov 28 '13 at 03:34
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(Too long for a comment.) Just to give the exact value of $(2)$ denoted as $A$ and its approximation denoted as $B$,

\begin{align} A = \frac{1}{e^{5\pi/6}\,\eta\big(\tfrac{2\,i}{5}\big)} = \frac{2^{19/8}(-1+5^{1/4})\,\pi^{3/4}}{e^{5\pi/6}\,(-1+\sqrt{5})^{3/2}\,\Gamma\big(\tfrac{1}{4}\big)} &\approx 0.0887758501511156259651860669\color{red}6\\[6pt] B =\Big(1-\frac{1}{5^{1/4}}\Big)^2\Big(\frac{1+\sqrt{5}}{4}\Big) &\approx 0.0887758501511156259651860669\color{red}2\end{align}

with the difference $A-B\,$ a mere $10^{-29}$. Why that is so seems to be still unexplained.

P.S. Other exact values of the Dedekind eta function $\eta(\tau)$ are in this post.

Tito Piezas III
  • 60,745
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    I think using Ramanujan's class invariant $G_{25}=\phi$ we get $\frac{1+\sqrt{5}}{2}=2^{-1/4}\frac{\eta(5i/2)^2}{\eta(5i)\eta(5i/4)}$ and we have an algebraic relation between $\eta(5i/2)$ and $\eta(2i/5)$. – Paramanand Singh Apr 10 '25 at 05:33
  • @ParamanandSingh I've edited the answer above to highlight the 29 decimal places that $A$ and $B$ share and which makes it look nicer. Perhaps you can expand your comment about $G_{25}$ as an answer? – Tito Piezas III Apr 10 '25 at 06:18
  • @ParamanandSingh Good to know and thanks. Comment has been deleted. – Tito Piezas III Apr 14 '25 at 02:58