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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
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2 Answers2

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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 $(2)$, we have the identity,

$$A = \frac{1}{e^{5\pi/6}\,\eta\big(\tfrac{2\,i}{5}\big)} = \frac{2^{19/8}(-1+5^{1/4})\,\pi^{3/4}}{e^{5\pi/6}\,(-1+\sqrt{5})^{3/2}\,\Gamma\big(\tfrac{1}{4}\big)} = 0.0887758\dots$$

which is approximated by,

$$B =\Big(1-\frac{1}{5^{1/4}}\Big)^2\Big(\frac{1+\sqrt{5}}{4}\Big)= 0.0887758\dots$$

and $A\approx B$ with the aforementioned difference of 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.

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Tito Piezas III
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