Super Golden Ratio
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08-07-2022, 07:55 AM
Post: #1
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Super Golden Ratio
Recently I stumbled upon this video by Micheal Penn: What is the super-golden ratio??
It is defined by this equation: \( \begin{align} \psi^3=\psi^2+1 \end{align} \) He solves it algebraically to find: \( \begin{align} \psi=\frac{1}{3}\left(1+\sqrt[3]{\frac{1}{2}\left(29-3\sqrt{93}\right)}+\sqrt[3]{\frac{1}{2}\left(29+3\sqrt{93}\right)}\right) \end{align} \) We can calculate a numerical approximation using Bernoulli's Method with the following program for the HP-42S: Code: 00 { 4-Byte Prgm } Initialise the stack with: CLST 1 And then hit repeatedly the R/S key: R/S R/S R/S … It produces the following sequence: 1 1 1 2 3 4 6 9 13 19 28 … This is also known as A000930: Narayana's cows sequence. The quotients of consecutive elements converge to the root and thus to \(\psi\): Code: 1 ÷ 1 = 1.00000000000 But this is the power iteration to calculate the greatest (in absolute value) eigenvalue of a diagonalizable matrix. To see this enter the following matrix with: 3 ENTER NEWMAT \( \begin{bmatrix} 1 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{bmatrix} \) Fill the stack using: ENTER ENTER ENTER Enter the initial vector: \( \begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix} \) 3 ENTER 1 NEWMAT 1 + And now repeatedly hit the multiply key: * * * … If you think you had enough, calculate the quotient of two consecutive elements, e.g. for: \( \begin{bmatrix} 58425 \\ 39865 \\ 27201 \\ \end{bmatrix} \) 58425 39865 ÷ 1.46557130315 (1.46557123188) This is a "Good place to stop". References You can find a description in Computational analysis with the HP-25 pocket calculator (Peter Henrici):
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08-07-2022, 07:56 AM
(This post was last modified: 08-07-2022 08:41 AM by Thomas Klemm.)
Post: #2
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RE: Super Golden Ratio
If you want some more do the same with the Plastic number or the Tribonacci numbers.
Or then use one of the many algorithms to accelerate the convergence of the original sequence. |
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08-08-2022, 04:07 PM
Post: #3
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RE: Super Golden Ratio
“Rabbit Math” – A Formula for the Magical Fibonacci Sequence
We use the same trick, for "Super" Fibonacci numbers generation function x/(1-x-x^3) = x + x^2 + x^3 + 2*x^4 + 3*x^5 + 4*x^6 + 6*x^7 + 9*x^8 + ... Binet-like direct formula for "Super" Fibonacci numbers, has this form: 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 XCAS> R := 1 ./ proot([-1,0,-1,1]); → [-0.232786+0.792552*i, -0.232786-0.792552*i, 1.46557] XCAS> K := proot([-1,0,3/31,1/31]); → [-0.208619-0.183825*i, -0.208619+0.183825*i, 0.417238] XCAS> [K*R.^29, K*R.^30, K*R.^31] → [27201.0, 39865.0, 58425.0] Both cubics are in depressed form, easy to get (K,R) exact form w = exp(i*2*pi/3) :; kb := (1/62 + sqrt(93)*3/1922)^(1/3) :; ka := +1/(31*kb) :; rb := (1/2 + sqrt(93)/18)^(1/3) :; ra := -1/(3*rb) :; K := [ka*w+kb/w, ka/w+kb*w, ka+kb] :; R := 1 ./ [ra*w+rb/w, ra/w+rb*w, ra+rb] :; approx(K,R) matched proot values. We just need to confirm K coefficients produce S(1) = S(2) = S(3) = 1, exactly. XCAS> S(n) := K * R.^n XCAS> simplify([S(1), S(2), S(3)]) After 31 seconds, we get back [1, 1, 1] |
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08-09-2022, 10:45 AM
(This post was last modified: 08-09-2022 11:08 AM by Gerson W. Barbosa.)
Post: #4
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RE: Super Golden Ratio
(08-07-2022 07:55 AM)Thomas Klemm Wrote: We can calculate a numerical approximation using Bernoulli's Method with the following program for the HP-42S: This will do for the 12 digits in the display: 67 ENTER X↑2 1/X + 11 1/X Y↑X Or, for more digits, (67 + 1/(67^2 + 2/(67 + 3/(2×67^2 + 11/(3×67)))))^(1/11) = 1.46557123187676802665673122475 The latter should take up more steps than the exact expression, though. P.S.: 300766 ENTER 33 1/X Y↑X is shorter for 12 digits and the approximation is slightly better. |
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08-12-2022, 05:46 PM
Post: #5
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RE: Super Golden Ratio
(08-08-2022 04:07 PM)Albert Chan Wrote: S(n) = [k1, k2, k3] * [r1, r2, r3] .^ n 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 } 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 } 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]); 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? |
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08-12-2022, 10:01 PM
Post: #6
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RE: Super Golden Ratio
(08-12-2022 05:46 PM)Thomas Klemm Wrote: We assume that \(\alpha\), \(\beta\) and \(\gamma\) are the roots of \(x^3=x^2+1\). Simpler approach is do partial fraction decomposition of generation function. S(n) formula is coefficient of x^n, with RHS geometric series in normalized form. \(\displaystyle \frac{x}{(1 - α x)(1- β x) (1 - γ x)} = \frac{u}{1 - α x} + \frac{v}{1- β x} + \frac{w}{1 - γ x} \) Mulitply both side by (1 - α x), then let x = 1/α, we solved u \(\displaystyle u = \frac{1/α}{(1 - β/α) (1 - γ/α)} = \frac{α}{(α-β)(α-γ)}\) \(\displaystyle \frac{1}{u} = α - (β+γ) + \frac{βγ}{α} = α - (1-α) + (α-1) = (3α - 2)\) \(\displaystyle \frac{1}{u} + \frac{1}{v} + \frac{1}{w} = (3α\!-\!2) + (3β\!-\!2) + (3γ\!-\!2) = 3 - 6 = -3 \) If we let x = 0, we get 0 = u + v + w \(\displaystyle \;⇒ \frac{1}{uv} + \frac{1}{uw} + \frac{1}{vw} = 0\) \(\displaystyle \frac{1}{uvw} = (3α\!-\!2)(3β\!-\!2)(3γ\!-\!2) = 27(αβγ) - 18(αβ\!+\!αγ\!+\!βγ) + 12(α\!+\!β\!+\!γ) - 8 = 27 + 12 - 8 = 31 \) \(\displaystyle ⇒ (x-\frac{1}{u})(x-\frac{1}{v})(x-\frac{1}{w}) = x^3 + 3x^2 - 31\) |
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08-12-2022, 10:59 PM
Post: #7
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RE: Super Golden Ratio
Another way to get cubic with roots = (roots of x^3-x^2-1) * 3 - 2
CAS> subst(x^3 - x^2 - 1, x = (y+2)/3) ((y+2)/3)^3 - ((y+2)/3)^2 - 1 CAS> numer(Ans) // y has roots 1/u, 1/v, 1/w y^3 + 3*y^2 - 31 |
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08-13-2022, 09:57 AM
Post: #8
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RE: Super Golden Ratio
Thanks a lot for your explanations.
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08-17-2022, 02:23 PM
Post: #9
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RE: Super Golden Ratio
(08-09-2022 10:45 AM)Gerson W. Barbosa Wrote: 300766 ENTER 33 1/X Y↑X is shorter for 12 digits and the approximation is slightly better. Let (α,β,γ) be roots of x^3 = x^2 + 1 We can build cubic with roots (-α^n, -β^n, -γ^n) n=1: x^3 + x^2 + 1 n=2: x^3 + x^2 - 2x + 1 n=3: x^3 + 4x^2 + 3x + 1 x^2 term coefficient = (α^+n + β^+n + γ^+n) x^1 term coefficient = (α^−n + β^−n + γ^−n), since αβγ = 1 x^+n = x^+(n-3) + x^+(n-1) x^−n = x^−(n-3) − x^−(n-2) lua> B, C = {1,1,4}, {0,-2,3} lua> for n=4,40 do B[n]=B[n-3]+B[n-1]; C[n]=C[n-3]-C[n-2] end lua> for n=30,36 do print(n, B[n], C[n]) end Code: 30 95545 406 For n=33, linear term coefficient is relatively tiny. Also, (α^n, β^n, γ^n) well separated. lua> surd(B[33], 33) -- estimated Psi 1.4655712318782408 -- lua> B[33]^2-2*C[33], C[33]^2-2*B[33] --Graeffe Root Squaring 90460186750 -601523 Free42: (90460186750)^(1/66) 1.465571231876768025024430123337594 Free42: (90460186750^2 + 2*601523)^(1/132) // "square" again 1.465571231876768026656731225219939 |
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08-20-2022, 05:10 PM
(This post was last modified: 08-21-2022 06:24 PM by Albert Chan.)
Post: #10
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RE: Super Golden Ratio
ψ^3 = (ψ^2 + 1)
ψ^4 = (ψ^2 + 1)*ψ = (ψ^2 + 1) + ψ = (1*ψ^2 + ψ + 1) ψ^5 = (ψ^2 + 1)*ψ^2 = (ψ^2 + ψ + 1) + ψ^2 = (2*ψ^2 + ψ + 1) ψ^6 = (ψ^2 + 1)^2 = (ψ^2 + ψ + 1) + 2*ψ^2 + 1 = (3*ψ^2 + ψ + 2) ... ψ^n = S(n-1)*ψ^2 + S(n-3)*ψ + S(n-2) ψ^(2n) = S(2n-1)*ψ^2 + S(2n-3)*ψ + S(2n-2) ψ^(3n) = S(3n-1)*ψ^2 + S(3n-3)*ψ + S(3n-2) ψ^(3n) = k1*ψ^(2n) + k2*ψ^n + k3 k1 = round(ψ^n), for n ≥ 6 → ψ ≈ k1^(1/n) k3 = (αβγ)^n = 1 For big n, S(3n-1)/S(3n-3) is better estimate for ψ^2, than S(2n-1)/S(2n-3). Thus, we expected k2 ≠ 0. But, there are exceptions, when n is small. n = 1: ψ^3 = ψ^2 + 1 // n=1 --> k2=0 n=11: S(3n-1)/S(3n-3) = 85626/39865 = (67*1278)/(67*595) = 1278/595 = S(2n-1)/S(2n-3) ψ^33 = 67*ψ^22 + 1 // n=11 --> k2=0 ψ^11 = 67 + 1/ψ^22 = 67 + 1/(67 + 1/ψ^22)^2 (08-09-2022 10:45 AM)Gerson W. Barbosa Wrote: Or, for more digits, This also explained why ψ^33 is very close to integer. With cubing of roots, "linear" term is still small. (-ψ^11)^3 + 67*(-ψ^11)^2 + 1 = 0 lua> b, c = 67, 0 lua> b*(b*b-3*c)+3, c*(c*c-3*b)+3 300766 3 (-ψ^33)^3 + 300766*(-ψ^33)^2 + 3*(-ψ^33) + 1 = 0 ψ^33 = 300766 - 3/ψ^33 + 1/ψ^66 ≈ 300766 - 3/300766 + 1/300766^2 ≈ 300765.9999900255 |
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