Supergolden Ratio is Unique Real Root of x^3 - x^2 - 1
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Definition
The supergolden ratio $\psi$ is the one real root of the cubic:
- $x^3 - x^2 - 1 = 0$
Its decimal expansion starts:
- $\eta = 1 \cdotp 46557 \, 12318 \, 76768 \ldots$
This sequence is A092526 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).
Proof
Part 1: Reduction
\(\ds x^3 - x^2 - 1\) | \(=\) | \(\ds 0\) | Given | |||||||||||
\(\ds \paren { u + \dfrac 1 3 }^3 - \paren { u + \dfrac 1 3 }^2 - 1\) | \(=\) | \(\ds 0\) | Substitution: $u = x - \dfrac b {3a}$ | |||||||||||
\(\ds u^3 - \dfrac 1 3 u - \dfrac {29} {27}\) | \(=\) | \(\ds 0\) | Expanding polynomials; trivial simplification | |||||||||||
\(\ds \paren { v + v^{-1} w }^3 - \dfrac 1 3 \paren { v + v^{-1} w } - \dfrac {29} {27}\) | \(=\) | \(\ds 0\) | Substitution: $u = v + \dfrac w v$ | |||||||||||
\(\ds \paren 1 \ \ \) | \(\ds v^3 \paren { v + v^{-1} w }^3 - \dfrac 1 3 v^3 \paren { v + v^{-1} w } - \dfrac {29} {27} v^3\) | \(=\) | \(\ds 0\) | Adding degrees to a polynomial set equal to zero does not produce non-zero roots | ||||||||||
\(\ds v^3 \paren { v^3 + 3vw + 3 v^{-1} w^2 + v^{-3} w^3 } - \dfrac 1 3 v^3 \paren { v + v^{-1} w } - \dfrac {29} {27} v^3\) | \(=\) | \(\ds 0\) | Expanding polynomials | |||||||||||
\(\ds v^6 + 3 v^4 w + 3 v^2 w^2 + w^3 - \dfrac 1 3 v^4 - \dfrac 1 3 v^2 w - \dfrac {29} {27} v^3\) | \(=\) | \(\ds 0\) | Distributive Property | |||||||||||
\(\ds v^6 + v^4 \paren { 3 w - \dfrac 1 3 } + v^2 \paren { 3 w^2 - \dfrac 1 3 w } - \dfrac {29} {27} v^3 + w^3\) | \(=\) | \(\ds 0\) | Rewriting in terms of $v$ | |||||||||||
\(\ds v^6 - \dfrac {29} {27} v^3 + \dfrac {1} {729}\) | \(=\) | \(\ds 0\) | Substituting $w = \dfrac 1 9$ (only degrees of $v$ divisible by $v^3$ from $\paren 1$ remain) | |||||||||||
\(\ds h^2 - \dfrac {29} {27} h + \dfrac {1} {729}\) | \(=\) | \(\ds 0\) | Substituting $h = v^3$ |
Part 2: Complete the square
\(\ds h^2 - \dfrac {29} {27} h + \dfrac {1} {729}\) | \(=\) | \(\ds 0\) | From reduction | |||||||||||
\(\ds h^2 - \dfrac {29} {27} h + \dfrac {841} {2916}\) | \(=\) | \(\ds \dfrac {837} {2916}\) | adding $\dfrac {837} {2916}$ to both sides | |||||||||||
\(\ds \paren { h - \dfrac {29} {54} }^2\) | \(=\) | \(\ds \dfrac {837} {2916}\) | $a^2 + 2ab + b^2 = \paren { a + b }^2$ | |||||||||||
\(\ds h - \dfrac {29} {54}\) | \(=\) | \(\ds ± \dfrac { \sqrt {93} } {18}\) | taking the square root of both sides | |||||||||||
\(\ds h\) | \(=\) | \(\ds \dfrac { 29 ± 3 \sqrt {93} } {54}\) | adding $\dfrac {29} {54}$ to both sides |
Part 3: Substitution
Initial substitution
\(\ds v^3\) | \(=\) | \(\ds \dfrac { 29 ± 3 \sqrt {93} } {54}\) | Substituting: $h = v^3$ | |||||||||||
\(\ds \Lambda\) | \(=\) | \(\ds 29 ± 3 \sqrt {93}\) | multiplying both sides by 54, seeing as we will soon do a substitution | |||||||||||
\(\ds \Lambda\) | \(=\) | \(\ds 29 + 3 \sqrt {93}\) | Sign rectification |
We make a sign rectification because this cascade works with only one substitution at a time.
We choose $+$ because we know $x^3 - x^2$ 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
\(\ds \dfrac 1 3 \paren { -1 }^{ \dfrac 2 3 } \paren {2}^{- \dfrac 1 3} {\Lambda}^{ \dfrac 1 3 } - \dfrac 1 3 \paren {2}^{\dfrac 1 3} \paren { - \Lambda }^{ - \dfrac 1 3 }\) | \(=\) | \(\ds u\) | From substitution | |||||||||||
\(\ds \dfrac 1 3 \paren { \sqrt [3] {-1} }^2 \paren {2}^{- \dfrac 1 3} {\Lambda}^{ \dfrac 1 3 } - \dfrac 1 3 \paren {2}^{\dfrac 1 3} \paren { - \Lambda }^{ - \dfrac 1 3 }\) | \(=\) | \(\ds u\) | Cube root factoring | |||||||||||
\(\ds \dfrac 1 3 \paren { -1 }^2 \paren {2}^{- \dfrac 1 3} {\Lambda}^{ \dfrac 1 3 } - \dfrac 1 3 \paren {2}^{\dfrac 1 3} \paren { - \Lambda }^{ - \dfrac 1 3 }\) | \(=\) | \(\ds u\) | taking the real cube root of $-1$ | |||||||||||
\(\ds \dfrac 1 3 \paren {2}^{- \dfrac 1 3} \paren {\Lambda}^{ \dfrac 1 3 } - \dfrac 1 3 \paren {2}^{\dfrac 1 3} \paren { - \Lambda }^{ - \dfrac 1 3 }\) | \(=\) | \(\ds u\) | taking the square of $-1$ | |||||||||||
\(\ds \dfrac 1 3 \paren {2}^{- \dfrac 1 3} \paren {\Lambda}^{ \dfrac 1 3 } - \dfrac 1 3 \paren {2}^{\dfrac 1 3} \paren { - \Lambda }^{ - \dfrac 1 3 } \paren { \overline \Lambda }^{ \dfrac 1 3 } \paren { \overline \Lambda }^{ - \dfrac 1 3 }\) | \(=\) | \(\ds u\) | multiplying the radicand by $\paren { \overline \Lambda }^{ \dfrac 1 3 } \paren { \overline \Lambda }^{ - \dfrac 1 3 }$ | |||||||||||
\(\ds \dfrac 1 3 \paren {2}^{- \dfrac 1 3} \paren {\Lambda}^{ \dfrac 1 3 } - \dfrac 1 3 \paren {2}^{\dfrac 1 3} \paren { - \dfrac 1 4 \overline \Lambda }^{ \dfrac 1 3 }\) | \(=\) | \(\ds u\) | $\paren { 29 + 3 \sqrt {93} } \paren { 29 - 3 \sqrt {93} } = 4$ | |||||||||||
\(\ds \dfrac 1 3 \paren {2}^{- \dfrac 1 3} \paren {\Lambda}^{ \dfrac 1 3 } + \dfrac 1 3 \paren {2}^{\dfrac 1 3} \paren { \dfrac 1 4 \overline \Lambda }^{ \dfrac 1 3 }\) | \(=\) | \(\ds u\) | Sign factoring | |||||||||||
\(\ds \dfrac 1 3 \paren {2}^{- \dfrac 1 3} {\Lambda}^{ \dfrac 1 3 } + \dfrac 1 3 \paren { \dfrac 1 4 }^{ \dfrac 1 3 } \paren {2}^{\dfrac 1 3} \paren { \overline \Lambda }^{ \dfrac 1 3 }\) | \(=\) | \(\ds u\) | Radical factoring | |||||||||||
\(\ds \dfrac 1 3 \paren {2}^{- \dfrac 1 3} {\Lambda}^{ \dfrac 1 3 } + \dfrac 1 3 \paren {2}^{- \dfrac 1 3} { \overline \Lambda }^{ \dfrac 1 3 }\) | \(=\) | \(\ds u\) | Radical factoring | |||||||||||
\(\ds \dfrac 1 3 \paren {\dfrac {\Lambda} 2}^{ \dfrac 1 3 } + \dfrac 1 3 \paren {\dfrac { \overline \Lambda } 2 }^{ \dfrac 1 3 }\) | \(=\) | \(\ds u\) | Radical factoring | |||||||||||
\(\ds \dfrac { \sqrt [3] { \dfrac {29 + 3 \sqrt {93} } {2} } + \sqrt [3] { \dfrac {29 - 3 \sqrt {93} } {2} } } 3\) | \(=\) | \(\ds u\) | Rewriting in radical notation | |||||||||||
\(\ds \dfrac { \sqrt [3] { \dfrac {29 + 3 \sqrt {93} } {2} } + \sqrt [3] { \dfrac {29 - 3 \sqrt {93} } {2} } } 3\) | \(=\) | \(\ds x - \dfrac 1 3\) | Substituting: $u = x - \dfrac b {3a}$ | |||||||||||
\(\ds \dfrac { 1 + \sqrt [3] { \dfrac {29 + 3 \sqrt {93} } {2} } + \sqrt [3] { \dfrac {29 - 3 \sqrt {93} } {2} } } 3\) | \(=\) | \(\ds x\) | adding $\dfrac 1 3$ to both sides |
Therefore, the supergolden ratio is equal to the preceding. It can be evaluated manually or with a calculator to be approx. $1.46557$.
$\blacksquare$