Supergolden Ratio is Unique Real Root of x^3 - x^2 - 1

== Definition ==
The supergolden ratio $\psi$ is the one real root of the cubic:
:$x^3 - x^2 - 1 = 0$
Its decimal expansion starts:
:$\eta = 1 \cdotp 46557 \, 12318 \, 76768 \ldots$
== Proof ==
=== Part 1: Reduction ===
=== Part 2: Complete the square ===
=== Part 3: Substitution ===
==== Initial substitution ====
We make a sign rectification because this cascade works with only one substitution at a time.
We choose $+$ because we know $x^3 - x^2$ has a zero root and a positive root, so $x^3 - x^2 - 1$ 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:
$v^3 = \dfrac \Lambda {54}$
$v = \dfrac {1} { 3 \sqrt [3] {2} } { \Lambda }^{ \dfrac 1 3 } = \dfrac 1 3 \paren {2}^{- \dfrac 1 3} \paren {\Lambda}^{\dfrac 1 3}$
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:
$v = - \dfrac 1 3 \paren {2}^{- \dfrac 1 3} \paren { - \Lambda }^{ \dfrac 1 3 }$
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:
$v = \dfrac 1 3 \paren { -1 }^{ \dfrac 2 3 } \paren {2}^{- \dfrac 1 3} { \Lambda }^{ \dfrac 1 3 }$
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 $w = \dfrac 1 9$.
$u = \dfrac 1 3 \paren {2}^{\dfrac 1 3} \paren {\Lambda}^{- \dfrac 1 3} + \dfrac 1 3 \paren {2}^{- \dfrac 1 3} \paren {\Lambda}^{\dfrac 1 3}$
$u = - \dfrac 1 3 \paren {2}^{\dfrac 1 3} \paren { - \Lambda }^{ - \dfrac 1 3 } + \dfrac 1 3 \paren { -1 }^{ \dfrac 2 3 } \paren {2}^{- \dfrac 1 3} \paren { \Lambda }^{ \dfrac 1 3 }$
$u = \dfrac 1 3 \paren { -1 }^{ \dfrac 2 3 } \paren {2}^{\dfrac 1 3} \paren { \Lambda }^{ - \dfrac 1 3 } - \dfrac 1 3 \paren {2}^{- \dfrac 1 3} \paren {- \Lambda }^{ \dfrac 1 3 }$
We can pick any of these three roots. The following will use the second of these three.
For reference, let ${ \overline \Lambda } = 29 - 3 \sqrt {93}$, the negative form.
=== Part 4: Lambda cubic cascade ===
Therefore, the supergolden ratio is equal to the preceding. It can be evaluated manually or with a calculator to be approx. $1.46557$.
Category:Cubic Equations