Tribonacci Constant is Unique Real Root of x^3-x^2-x-1

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Theorem

The Tribonacci constant $\eta$ is the one real root of the cubic:

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


Its decimal expansion starts:

$\eta = 1 \cdotp 83928 \, 67552 \, 1416 \ldots$

This sequence is A058265 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


Proof


Although this article appears correct, it's inelegant. There has to be a better way of doing it.
In particular: This is basically just following the workings of Cardano's Formula for a specific set of numbers. Do we really need to do this? Recommend it be replaced by an invocation of that formula.
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Part 1: Reduction

\(\ds x^3 - x^2 - x - 1\) \(=\) \(\ds 0\) Given
\(\ds \paren { u + \dfrac 1 3 }^3 - \paren { u + \dfrac 1 3 }^2 - u - \dfrac 4 3\) \(=\) \(\ds 0\) Substitution: $u = x - \dfrac b {3a}$
\(\ds u^3 - \dfrac 4 3 u - \dfrac {38} {27}\) \(=\) \(\ds 0\) Expanding polynomials; trivial simplification
\(\ds \paren { v + v^{-1} w }^3 - \dfrac 4 3 \paren { v + v^{-1} w } - \dfrac {38} {27}\) \(=\) \(\ds 0\) Substitution: $u = v + \dfrac w v$
\(\ds \paren 1 \ \ \) \(\ds v^3 \paren { v + v^{-1} w }^3 - \dfrac 4 3 v^3 \paren { v + v^{-1} w } - \dfrac {38} {27} v^3\) \(=\) \(\ds 0\) Adding degrees to a polynomial set equal to zero does not produce non-zero roots
\(\ds v^3 \paren { v^3 + 3vw + 3 v^{-1} w^2 + v^{-3} w^3 } - \dfrac 4 3 v^3 \paren { v + v^{-1} w } - \dfrac {38} {27} v^3\) \(=\) \(\ds 0\) Expanding polynomials
\(\ds v^3 + 3 v^4 w + 3 v^2 w^2 + w^3 - \dfrac 4 3 v^4 - \dfrac 4 3 v^2 w - \dfrac {38} {27} v^3\) \(=\) \(\ds 0\) Distributive Property
\(\ds v^6 + v^4 \paren { 3 w - \dfrac 4 3 } + v^2 \paren { 3 w^2 - \dfrac 4 3 w } - \dfrac {38} {27} v^3 + w^3\) \(=\) \(\ds 0\) Rewriting in terms of $v$
\(\ds v^6 - \dfrac {38} {27} v^3 + \dfrac {64} {729}\) \(=\) \(\ds 0\) Substituting $w = \dfrac 4 9$ (only degrees of $v$ divisible by $v^3$ from $\paren 1$ remain)
\(\ds h^2 - \dfrac {38} {27} h + \dfrac {64} {729}\) \(=\) \(\ds 0\) Substituting $h = v^3$


Part 2: Complete the square

\(\ds h^2 - \dfrac {38} {27} h + \dfrac {64} {729}\) \(=\) \(\ds 0\) From reduction
\(\ds h^2 - \dfrac {38} {27} h + \dfrac {361} {729}\) \(=\) \(\ds \dfrac {297} {729}\) adding $\dfrac {297} {729}$ to both sides
\(\ds \paren { h - \dfrac {19} {27} }^2\) \(=\) \(\ds \dfrac {297} {729}\) $a^2 + 2ab + b^2 = \paren { a + b }^2$
\(\ds h - \dfrac {19} {27}\) \(=\) \(\ds ± \dfrac { 3 \sqrt {33} } {27}\) taking the square root of both sides
\(\ds h\) \(=\) \(\ds \dfrac { 19 ± 3 \sqrt {33} } {27}\) adding $\dfrac {19} {27}$ to both sides


Part 3: Substitution

Initial substitution

\(\ds v^3\) \(=\) \(\ds \dfrac { 19 ± 3 \sqrt {33} } {27}\) Substituting: $h = v^3$
\(\ds \Lambda\) \(=\) \(\ds 19 ± 3 \sqrt {33}\) adding $\dfrac {19} {27}$ to both sides
\(\ds \Lambda\) \(=\) \(\ds 19 + 3 \sqrt {33}\) Sign rectification

We make a sign rectification because this cascade works with only one substitution at a time.

We choose $+$ because we know $x^3 - x^2 - x$ has a zero root and a positive root, so $x^3 - x^2 - x - 1$ can only have a positive root.

The cascade will account for the $-$ solution anyway.


Lambda substitution

This next part of the substitution occurs because of the following logic:

$P1:$ To take a cube root of a real number, there must be three possible solutions (one real, two complex).

$P2:$ This polynomial is being restricted to the real xy-plane, i.e. no complex roots are allowed.

$C1:$ To take the cube root of this real number, it needs to vary by signs as the two complex would.

$P3:$ A complex number has its own arguments with which sign works, i.e. $+i$ vs. $-i$, which does not apply on the real number line.

$C2:$ All combinations of signs have to be represented, and there must be three.

$P4:$ Since a sign has two states, $+$ and $-$, there must be at least one other entity able to be modified with a sign.

$C3:$ To take the cube root of this real number, two parts of it must be able to be modified with a sign together exactly thrice.

We can see that the first cube root is trivially determined:

$v^3 = \dfrac \Lambda {27}$

$v = \dfrac 1 3 { \Lambda }^{ \dfrac 1 3 }$

The two parts we can modify are $\dfrac 1 3$ and $\Lambda$. We already have $+$ for the first and $+$ for the second.

We know $-1 × -1 = +1$, so we can have $-$ for the first and $-$ for the second:

$v = - \dfrac 1 3 \paren { - \Lambda }^{ \dfrac 1 3 }$

There must be a third way. We can have both negative signs on the first part or on the second part (as a cube root). It's possible to write it as both at the same time if it is in the form of a third part, so that it can multiply by either:

$v = \dfrac 1 3 \paren { -1 }^{ \dfrac 2 3 } { \Lambda }^{ \dfrac 1 3 }$

Any of these three roots are able to be used in the cascade.


Reverse substitution

Substitute for $u = v + \dfrac w v$ in each of the three roots, where $w = \dfrac 4 9$.

$u = \dfrac 4 3 \paren { \Lambda }^{ - \dfrac 1 3 } + \dfrac 1 3 \paren { \Lambda }^{ \dfrac 1 3 }$

$u = - \dfrac 4 3 \paren { - \Lambda }^{ - \dfrac 1 3 } + \dfrac 1 3 \paren { -1 }^{ \dfrac 2 3 } \paren { \Lambda }^{ \dfrac 1 3 }$

$u = \dfrac 4 3 \paren { -1 }^{ \dfrac 2 3 } \paren { \Lambda }^{ - \dfrac 1 3 } - \dfrac 1 3 \paren { - \Lambda }^{ \dfrac 1 3 }$

We can pick any of these three roots. The following will use the second of these three.

For reference, let ${ \overline \Lambda } = 19 - 3 \sqrt {33}$, the negative form.


Part 4: Lambda cubic cascade

\(\ds \dfrac 1 3 \paren { -1 }^{ \dfrac 2 3 } {\Lambda}^{ \dfrac 1 3 } - \dfrac 4 3 \paren { - \Lambda }^{ - \dfrac 1 3 }\) \(=\) \(\ds u\) From substitution
\(\ds \dfrac 1 3 \paren { \sqrt [3] {-1} }^2 {\Lambda}^{ \dfrac 1 3 } - \dfrac 4 3 \paren { - \Lambda }^{ - \dfrac 1 3 }\) \(=\) \(\ds u\) Cube root factoring
\(\ds \dfrac 1 3 \paren { -1 }^2 {\Lambda}^{ \dfrac 1 3 } - \dfrac 4 3 \paren { - \Lambda }^{ - \dfrac 1 3 }\) \(=\) \(\ds u\) taking the real cube root of $-1$
\(\ds \dfrac 1 3 \paren {\Lambda}^{ \dfrac 1 3 } - \dfrac 4 3 \paren { - \Lambda }^{ - \dfrac 1 3 }\) \(=\) \(\ds u\) taking the square of $-1$
\(\ds \dfrac 1 3 \paren {\Lambda}^{ \dfrac 1 3 } - \dfrac 4 3 \paren { - \Lambda }^{ - \dfrac 1 3 } \paren { \overline \Lambda }^{ \dfrac 1 3 } \paren { \overline \Lambda }^{ - \dfrac 1 3 }\) \(=\) \(\ds u\) multiplying the radicand by $\paren { \overline \Lambda }^{ \dfrac 1 3 } \paren { \overline \Lambda }^{ - \dfrac 1 3 }$
\(\ds \dfrac 1 3 \paren {\Lambda}^{ \dfrac 1 3 } - \dfrac 4 3 \paren { - \dfrac 1 {64} \overline \Lambda }^{ \dfrac 1 3 }\) \(=\) \(\ds u\) $\paren { 19 + 3 \sqrt {33} } \paren { 19 - 3 \sqrt {33} } = 64$
\(\ds \dfrac 1 3 \paren {\Lambda}^{ \dfrac 1 3 } + \dfrac 4 3 \paren { \dfrac 1 {64} \overline \Lambda }^{ \dfrac 1 3 }\) \(=\) \(\ds u\) Sign factoring
\(\ds \dfrac 1 3 {\Lambda}^{ \dfrac 1 3 } + \dfrac 4 3 \paren { \dfrac 1 {64} }^{ \dfrac 1 3 } \paren { \overline \Lambda }^{ \dfrac 1 3 }\) \(=\) \(\ds u\) Radical factoring
\(\ds \dfrac 1 3 {\Lambda}^{ \dfrac 1 3 } + \dfrac 4 3 \paren { \dfrac 1 4 } { \overline \Lambda }^{ \dfrac 1 3 }\) \(=\) \(\ds u\) taking the cube root of $\dfrac 1 {64}$
\(\ds \dfrac 1 3 {\Lambda}^{ \dfrac 1 3 } + \dfrac 1 3 { \overline \Lambda }^{ \dfrac 1 3 }\) \(=\) \(\ds u\) $\dfrac 4 3 × \dfrac 1 4 = \dfrac 1 3$
\(\ds \dfrac { \sqrt [3] { 19 + 3 \sqrt {33} } + \sqrt [3] { 19 - 3 \sqrt {33} } } 3\) \(=\) \(\ds u\) Rewriting in radical notation
\(\ds \dfrac { \sqrt [3] { 19 + 3 \sqrt {33} } + \sqrt [3] { 19 - 3 \sqrt {33} } } 3\) \(=\) \(\ds x - \dfrac 1 3\) Substituting: $u = x - \dfrac b {3a}$
\(\ds \dfrac { 1 + \sqrt [3] { 19 + 3 \sqrt {33} } + \sqrt [3] { 19 - 3 \sqrt {33} } } 3\) \(=\) \(\ds x\) adding $\dfrac 1 3$ to both sides

Therefore, the Tribonacci constant is equal to the preceding.

It can be evaluated manually or with a calculator to be approximately $1.83929$.

$\blacksquare$

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