## Posts Tagged ‘Platonic solids’

### The Tremendous Tribonacci constant

The tribonacci constant is the real root of the cubic equation,

$T^3-T^2-T-1 = 0$

and is the limiting ratio of the tribonacci numbers = {0, 1, 1, 2, 4, 7, 13, 24, …} where each term is the sum of the previous three, analogous to the Fibonacci numbers.  Let $d = 11$, then,

$T = \frac{1}{3} +\frac{1}{3}\big(19+3\sqrt{3d}\big)^{1/3} + \frac{1}{3}\big(19-3\sqrt{3d}\big)^{1/3} = 1.839286\dots$

We’ll see that the tribonacci constant is connected to the complete elliptic integral of the first kind $K(k_{11})$.  But first, given the golden ratio’s infinite radical representation,

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

then T also has the beautiful infinite radical,

$\frac{1}{T-1} = \sqrt[3]{\frac{1}{2}+\sqrt[3]{\frac{1}{2}+\sqrt[3]{\frac{1}{2}+\sqrt[3]{\frac{1}{2}+\dots}}}} = 1.191487\dots$

as well as a continued fraction,

$\big(\frac{T}{T+1}\big) \big(e^{\frac{\pi\sqrt{11}}{24}}\big) = 1 + \cfrac{q}{1-q + \cfrac{q^3-q^2}{1+\cfrac{q^5-q^3}{1+\cfrac{q^7-q^4}{1+\ddots}}}} = 0.9999701\dots$

where q is the negative real number,

$q = \frac{-1}{e^{\pi \sqrt{11}}}$

Recall that at elliptic singular values, the complete elliptic integral of the first kind K(k) satisfies the equation,

$\frac{K'(k_d)}{K(k_d)} = \sqrt{d}$

or, in the syntax of Mathematica,

$\frac{EllipticK[1-ModularLambda[\sqrt{-d}]]}{EllipticK[ModularLamda[\sqrt{-d}]]} = \sqrt{d}$

Interestingly, we can express both $k_{11} = 0.000477\dots$ and $K(k_{11}) = 1.57098\dots$ in terms of the tribonacci constant as,

$k_{11} = \frac{1}{4}\left(2-\sqrt{\frac{2v+7}{2v-7}}\,\right) = 0.000477\dots$

where,

$v = T+4$

and,

$K(k_{11}) = \left(\frac{T+1}{T}\right)^2\, \frac{1}{11^{1/4}\, (4\pi)^{2}} \, \Gamma(\tfrac{1}{11}) \Gamma(\tfrac{3}{11}) \Gamma(\tfrac{4}{11}) \Gamma(\tfrac{5}{11}) \Gamma(\tfrac{9}{11})$

where $\Gamma(n)$ is the gamma function, as well as the infinite series,

$K(k_{11}) = \left(\frac{T+1}{T}\right)^2 \frac{\pi}{32^{1/4}} \sqrt{\frac{1}{4} \sum_{n=0}^\infty \frac{(6n)!}{(3n)!n!^3} \,\frac{1}{(-32)^{3n}} }$

With a slight tweak of the formula, we instead get,

$\frac{1}{4\pi} = \frac{1}{32^{3/2}} \sum_{n=0}^\infty \frac{(6n)!}{(3n)!n!^3}\, \frac{154n+15}{(-32)^{3n}}$

Finally, saving the best for last, given the snub cube, an Archimedean solid,

then the Cartesian coordinates for its vertices are all the even and odd permutations of,

{± 1,  ± 1/T,  ±}

with an even and odd number of plus signs, respectively, similar to how, for the vertices of the dodecahedron — a Platonic solid —  one can use the golden ratio.

For more about the tribonacci constant, and the equally fascinating plastic constant,  kindly refer to “A Tale of Four Constants “.