Super Golden Ratio

[Thomas Klemm]

(08-08-2022 04:07 PM)Albert Chan Wrote:  S(n) = [k1, k2, k3] * [r1, r2, r3] .^ n

[r1, r2, r3] are reciprocal roots of 1-x-x^3 = 0 (or, roots of x^3 = x^2+1)
[k1, k2, k3] are constants to be solved, so that S(1) = S(2) = S(3) = 1

We assume that \(\alpha\), \(\beta\) and \(\gamma\) are the roots of \(x^3=x^2+1\).
Thus we have:

\(
\begin{align}
\alpha + \beta + \gamma &= 1 \\
\alpha \cdot \beta \cdot \gamma &= 1 \\
\end{align}
\)

For the direct formula \(S(n) = u \cdot \alpha^n + v \cdot \beta^n + w \cdot \gamma^n\) the constants \(u\), \(v\) and \(w\) are solved so that: \(S(1) = S(2) = S(3) = 1\)

This leads to:

\(
\begin{bmatrix}
\alpha & \beta & \gamma \\
\alpha^2 & \beta^2 & \gamma^2 \\
\alpha^3 & \beta^3 & \gamma^3 \\
\end{bmatrix}
\begin{bmatrix}
u \\
v \\
w \\
\end{bmatrix}
=
\begin{bmatrix}
1 \\
1 \\
1 \\
\end{bmatrix}
\)

This can be solved for \(u\) to get:

\(
\begin{align}
u
&= \frac{(1-\beta)(1-\gamma)}{\alpha(\alpha-\beta)(\alpha-\gamma)} \\
&= \frac{1 - (\beta+\gamma) + \beta \gamma}{a[\alpha^2 - (\beta+\gamma)\alpha + \beta \gamma]} \\
&= \frac{1 - (1-\alpha) + \frac{1}{\alpha}}{\alpha[\alpha^2 - (1-\alpha)\alpha + \frac{1}{\alpha}]} \\
&= \frac{\alpha + \frac{1}{\alpha}}{\alpha[2\alpha^2 - \alpha + \frac{1}{\alpha}]} \\
&= \frac{1+ \frac{1}{\alpha^2}}{2\alpha^2 - \alpha + \frac{1}{\alpha}} \\
&= \frac{1+ \frac{1}{\alpha^2}}{2\alpha^2 - 2\alpha + \alpha +\frac{1}{\alpha}} \\
&= \frac{\alpha}{3\alpha^2 - 2\alpha} \\
&= \frac{1}{3\alpha - 2}
\end{align}
\)

And similarly for \(v\) and \(w\).
We consider \(\alpha = \psi\), i.e. the real solution which is dominant.

The following formula can be used to calculate \(\psi\):

\(
\begin{align}
\psi ={\frac {2}{3}}\cosh {\left({\tfrac {\cosh ^{-1}\left({\frac {29}{2}}\right)}{3}}\right)}+{\frac {1}{3}}
\end{align}
\)

This allows to calculate both \(\alpha\) in R00 and \(\frac{1}{u}\) in R01:

Code:

00 { 30-Byte Prgm }
01 29
02 2
03 ÷
04 ACOSH
05 3
06 ÷
07 COSH
08 2
09 ×
10 1
11 +
12 RCL ST X
13 3
14 ÷
15 STO 00
16 X<>Y
17 2
18 -
19 STO 01
20 END

R00: 1.465571231876768026656731225219939
R01: 2.396713695630304079970193675659816

We cheat a little bit and use only the dominant root and round the result to the next integer:

Code:

00 { 12-Byte Prgm }
01 RCL 00
02 X<>Y
03 Y↑X
04 RCL÷ 01
05 0.5
06 +
07 IP
08 END

Examples

1 R/S
1

2 R/S
1

3 R/S
1

4 R/S
2

5 R/S
3

6 R/S
4

7 R/S
6

29 R/S
27201

30 R/S
39865

31 R/S
58425

---

(08-08-2022 04:07 PM)Albert Chan Wrote:  XCAS> K := proot([-1,0,3/31,1/31]);
→ [-0.208619-0.183825*i, -0.208619+0.183825*i, 0.417238]

We can verify that indeed: \(31u^3=3u+1\)

\(
\begin{align}
\frac{3}{u^2} + \frac{1}{u^3}
&= 3(3\alpha-2)^2 + (3\alpha-2)^3 \\
&= (3\alpha-2)^2(3\alpha+1) \\
&= (9\alpha^2-12\alpha+4)(3\alpha+1) \\
&= 27\alpha^3 - 27\alpha^2 + 4 \\
&= 27(\alpha^3 - \alpha^2) + 4 \\
&= 27 + 4 \\
&= 31
\end{align}
\)

How did you come up with this equation?