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13 Apr

The Tremendous Tribonacci constant

Posted April 13, 2012 by tpiezas in algebra, calculus, equations, geometry, mathematics. Tagged: , , .

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,  ±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 “.

2 responses to this post.

  1. Posted by Tribonacci numbers, Padovan sequence, and more (Part 1) « tpiezas on May 21, 2012 at 5:41 pm

    […] already written about the tribonacci constant before.  But I want to include how Lin found that powers of t can be expressed in terms of those three […]

    Reply

  2. Posted by Alex on July 14, 2015 at 1:22 pm

    The intersection (X,Y) of the Bernoulli lemniscate and the line y = 1 – x.
    defines the lemniscate cotangent = (X/Y) = T
    where X = T/(T + 1)

    Reply

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