hbelabs.com

## Regular pentagon, lowest Dirichlet eigenvalue of Laplacian, accurate to one thousand (1000) digits

Robert S Jones
rsjones7 at yahoo dot com
July 1, 2015
I now have this eigenvalue to 1,500 digits (see OEIS A263202), but this page is still useful.
The calculation took 2.7 days on a laptop (quad core i7, 8 threads, 16 GB RAM). The largest matrix was $1145\times 1145$, i.e., $N=1145$ and the relative gap of the bound is $\epsilon\approx 2.76\times 10^{-1003}$, which means that $D\approx 1002.6$, slightly more than $1000$, and the global convergence rate was quite good at $\rho\approx D/N=1/1.14$. I did find it necessary to use a relatively high precision using "default(realprecision,round(1.7*N));" (pari/gp command), which for the largest $N$-value is $1947$ digits of precision. At that highest $N$-value, the matrix elements for each matrix took almost 3.5 hours to calculate, and another hour to calculate its determinant. It finished on July 1, 2015 at 10AM CST.

The answer is presented for a unit-edged pentagon, a pentagon with area $\pi$, and a pentagon inscribed in a unit-radius circle. The first and last seem popular, whereas the $\pi$-area version is interesting because it has the same area as the unit-radius circle and the eigenvalues within the equilateral triangle and the square with that area are relatively simple multiples of $\pi$ (see below).

In case you are wondering about the nature of these numbers, the canonical continued fraction representation, that reproduces at least the first 1000 digits, is also given. (I put boxes around the appropriate numbers to help clarify the representation.) For the really curious, the area-$\pi$ eigenvalue divided by $\pi$ is also given, since the eigenvalues of the equilateral triangle and square with that area are $4\pi/\sqrt{3}$ and $2\pi$, respectively.

The following results all represent the same calculation. If the unit-area regular pentagon has eigenvalue $\widehat{\lambda}$, the relationship is $\lambda=\widehat{\lambda}/A$ where $\lambda$ is the eigenvalue for area-$A$ regular pentagon. $$\lambda=\left\{\begin{array}{ll}\displaystyle\frac{4~\widehat{\lambda}}{5}\tan(\pi/5) & \mbox{~~~~~~~~[unit-edged regular pentagon]}\\ \displaystyle \frac{\widehat{\lambda}}{\pi} & \mbox{~~~~~~~~[\pi-area regular pentagon]}\\ \displaystyle \frac{\widehat{\lambda}}{5\sin(\pi/5)\cos(\pi/5)} & \mbox{~~~~~~~~[regular pentagon inscribed in unit radius circle]} \end{array}\right.$$

### Unit-Edged Regular Pentagon

The eigenvalue bound is $$\lambda=\underline{10.9964...15728}~871~_{160}^{464}$$ where the first thousand digits (underlined) are

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The continued fraction representation is $$\lambda=\fbox{10}+\frac{1}{\fbox{1}+\frac{1}{\fbox{278}+\frac{1}{\fbox{1}+\frac{1}{...}}}}$$
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### Regular Pentagon with Area $\pi$

The eigenvalue bound is $$\lambda=\underline{6.0221...701710}~08~_{073}^{240}$$ where the first thousand digits (underlined) are

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The continued fraction representation is $$\lambda=\fbox{6}+ \frac{1}{\fbox{45}+\frac{1}{\fbox{5}+\frac{1}{\fbox{1}+\frac{1}{...}}}}$$
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Keep in mind that for this polygon area, the equilateral triangle has $\lambda/\pi=4/\sqrt{3}=[2,3,4,3,4,3,4,...]\approx 2.309$, and the square has $\lambda/\pi=2=[2]$. Thus, for the triangle, this ratio is an algebraic (quadratic) irrational, and for the square, a rational (whole number). Perhaps it is interesting to see what the same number is for the regular pentagon.

For the regular pentagon, dividing the eigenvalue by $\pi$ yields $$\frac{\lambda}{\pi}=\underline{1.9169...410024}~22~_{793}^{847}$$ where the first thousand digits (underlined) are

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The continued fraction representation is $$\frac{\lambda}{\pi}= \fbox{1} + \frac{1}{\fbox{1}+\frac{1}{\fbox{11}+\frac{1}{\fbox{28}+\frac{1}{...}}}}$$
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To this precision, that number ($\lambda/\pi$) appears to be a transcendental irrational number since I can't find a repeating pattern. Indeed, I also don't see any simple, repeating pattern for any of the continued fractions on this page.

### Regular Pentagon Inscribed in Unit-Radius Circle

The eigenvalue bound is $$\lambda=\underline{7.9570...491348}~56~_{067}^{288}$$ where the first thousand digits (underlined) are

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The continued fraction representation is $$\lambda=\fbox{7} + \frac{1}{\fbox{1}+\frac{1}{\fbox{22}+\frac{1}{\fbox{3}+\frac{1}{...}}}}$$
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Note: In the future, perhaps I'll calculate the second eigenvalue to such precision and examine its ratio with this result. To sixty correctly rounded digits, the second, unit-edged regular pentagon Dirichlet eigenvalue is 27.7861978458853077069478632309749626234961221128315720050846, so to that precision, the ratio is 2.52683872972705709678237710631586809062753253260916261084181. This too does not have a simple repeating pattern in its continued fraction representation, [2, 1, 1, 8, 1, 4, 2, 2, 16, 3, 3, 1, 3, 1, 1, 2, 1, 1, 3, 1, 3, 3, 5, 1, 13, 1, 4, 1, 1, 13, 4, 3, 1, 4, 1, 1, 6, 5, 9, 1, 13, 2, 15, 1, 2, 3, 3, 1, 4, 9, 2, 14, 1, 4, 1, 7, 1, 1, 11, 1, 4, 5, 4, 1, 1, 1, 24, 8, 2, 1, 1, 1, 2, 52, 488, 2, 1, 3, 1, 1, 2, 56, 1, 1, 5, 1, 7, 6, 1, 19, 1, 2, 1, 1, 5, 1, 2, 1, 1, 9, 1, 1, 4, 2, 1, 2, 1, 13, 2, 2, 10, 2, 35, 19, 32]. Getting 1000 digits in that second eigenvalue will be a bit more difficult.