The Heegner Polynomials

Question

What is special about $x^3- 6 x^2 + 4 x -2$? The 24th power of the real root - 24 is curiously close to two other numbers, one being the Ramanujan constant.

There are more of these polynomials associated with the Heegner numbers.

$poly_1 = x^2 -2$. The first root space has the Pythagoras constant, Silver ratio, vertices of an octagon, A4 paper, Ammann tile and the curiosity below, with numbers representing powers of $\sqrt2$.

$poly_2 = x^3 - 2x^2 +2x-2$. The second root space is part of tribonacci space $T = t^3 - x^2 -x-1$, notable for the snub cube. Roots $T_n$ as ${T_n}^2 - T_n$ are the roots of $poly_2$. The polynomial roots $1+2 T_n -{T_n}^2$, also in the same root space, become de Weger's example, the second best known algebraic solution for the ABC Conjecture.

$poly_3 = x^3 - 2x-2$. This root space builds the 12 point Heilbronn solution and the 12 disk covering solution. With $r$ as the real root, the circles have radius $\sqrt{(1,r,r^2)\cdot(-3,0,1)}$, with two centers on the $x$-axis at $\sqrt{(1,r,r^2)\cdot(-7,4,0)}$ and $\sqrt{(1,r,r^2)\cdot(-1,2,-1)}$.

Can anyone find amazing properties for the root spaces of the last three polynomials?

These are related to New Substitution Tilings Using 2, φ, ψ, χ, ρ.

Answer 1

Here is how to derive the approximation $e^{\pi\sqrt{163}}\approx x^{24}-24$.

First recall how the approximation $e^{\pi\sqrt{163}}\approx640320^3+744$ is proved. The $j$ function has $q$-expansion $$j(\tau)=\frac{1}{q}+744+196884q+21493760q^2+\cdots.$$ If we set $\tau=\frac{1}{2}(1+\sqrt{-163})$ and $q=e^{2\pi i\tau}=-e^{-\pi\sqrt{163}}$, then we have $$-640320^3=j(\tau)=-e^{\pi\sqrt{163}}+744-\frac{196884}{e^{\pi\sqrt{163}}}+\frac{21493760}{e^{2\pi\sqrt{163}}}+\cdots.$$ This gives the approximation $$e^{\pi\sqrt{163}}\approx640320^3+744.$$

For the approximation $e^{\pi\sqrt{163}}\approx x^{24}-24$, we will use the Weber modular function $$\mathfrak f(\tau)=q^{-1/48}\prod_{n=1}^\infty(1+q^{n-1/2}).$$ The $24$th power of $\mathfrak f(\tau)$ has $q$-expansion $$\mathfrak f(\tau)^{24}=q^{-1/2}\prod_{n=1}^\infty(1+q^{n-1/2})^{24}=q^{-1/2}+24+276q^{1/2}+2048q+11202q^{3/2}+\cdots.$$ If we set $\tau=\sqrt{-163}$ and $q=e^{2\pi i\tau}=e^{-2\pi\sqrt{163}}$, then we have $$\mathfrak f(\sqrt{-163})^{24}=e^{\pi\sqrt{163}}+24+\frac{276}{e^{\pi\sqrt{163}}}+\frac{2048}{e^{2\pi\sqrt{163}}}+\frac{11202}{e^{3\pi\sqrt{163}}}+\cdots.$$ This gives the approximation $$e^{\pi\sqrt{163}}\approx\mathfrak f(\sqrt{-163})^{24}-24.$$ The improved accuracy of this approximation is due to the fact that the coefficient $276$ is smaller than the coefficient $196884$.

The only remaining mystery is why $\mathfrak f(\sqrt{-163})$ is the real root of the cubic polynomial $x^3-6x^2+4x-2$. One reference for this is the paper "On the Singular Values of Weber Modular Functions" by Yui and Zagier. From section 6 of the paper, we learn that there is a cubic polynomial \begin{align*} x^3-2\lambda x^2+2\mu x-2&=\left(x-\mathfrak f(\sqrt{-163})\right)\left(x-\frac{\sqrt2}{\mathfrak f\left(\frac{15+\sqrt{-163}}2\right)}\right)\left(x-\frac{\sqrt2}{\mathfrak f\left(\frac{-15+\sqrt{-163}}2\right)}\right)\\ &\approx(x-5.3186)(x-(0.3407+0.5099i))(x-(0.3407-0.5099i)) \end{align*} with $\lambda,\mu\in\mathbb{Z}$ (this uses the fact that $\mathbb{Q}(\sqrt{-163})$ has class number one). Then expanding the right hand side shows that $\lambda=3$ and $\mu=2$ as desired.

Answer 2

When dealing with exotic connections, sometimes it is helpful to take inspiration from Ramanujan. Thus, we'll bring in q-continued fractions, pi formulas, nested radicals, and sequences.

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I. q-Continued fractions

Let $q = -e^{-\pi\sqrt{11}}.\,$ Then,

$$\frac{1}{T}+1 = \cfrac{(-q)^{-1/24}} {1 + \cfrac{q} {1-q + \cfrac{q^3-q^2} {1 + \cfrac{q^5-q^3} {1 + \cfrac{q^7-q^4} {1+\ddots}}}}}$$

where $T$ is the tribonacci constant. This is reminiscent of Ramanujan's cfracs which involve the golden ratio. And not surprisingly, the number $24$ appears again.

In general, for $q = -e^{-\pi\sqrt{n}}$ and positive integer $n$, then the cfrac is a radical usually of high degree for large $n$. However, for $n = (11,19,43,67,163),$

$$x_n=\cfrac{(-q)^{-1/24}} {1 + \cfrac{q} {1-q + \cfrac{q^3-q^2} {1 + \cfrac{q^5-q^3} {1 + \cfrac{q^7-q^4} {1+\ddots}}}}}$$

then $x_n$ is just the real root of the five cubics mentioned by Pegg, namely $x^3-2x^2+2x-2 = 0$, and the others.

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II. Pi formulas

Given $v = x_{11} = \frac{1}{T}+1,$ where $T$ is still the tribonacci constant. Then,

$$\frac{1}{\pi} = \frac12\sum_{k=0}^\infty \frac{(2k)!^3}{k!^6}\frac{(2v-1)(4v-3)k+(2v-1)(v-1)-\frac12}{v^{24k}}$$

Is that our friend the number $24$ again? In fact, similar pi formulas can be found using $x_n$ for $n = (11,19,43,67,163).$

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III. Nested radicals

We have the well-known,

$$\phi =\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$$

with golden ratio $\phi$. The tribonacci version is,

$$\frac1{T-1}=\sqrt[3]{\frac12+\sqrt[3]{\frac12+\sqrt[3]{\frac12+\sqrt[3]{\frac12+\dots}}}}$$

Again, one can do the same for all cubic roots, but they may not be so pretty.

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IV. Sequences

To avoid redundancy, more details can be found in this post.