08-07-2022, 07:55 AM
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:
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 \(\psi\):
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):
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 }
01 ENTER
02 RCL+ ST T
03 END
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
2 ÷ 1 = 2.00000000000
3 ÷ 2 = 1.50000000000
4 ÷ 3 = 1.33333333333
6 ÷ 4 = 1.50000000000
9 ÷ 6 = 1.50000000000
13 ÷ 9 = 1.44444444444
19 ÷ 13 = 1.46153846154
28 ÷ 19 = 1.47368421053
41 ÷ 28 = 1.46428571429
60 ÷ 41 = 1.46341463415
88 ÷ 60 = 1.46666666667
129 ÷ 88 = 1.46590909091
189 ÷ 129 = 1.46511627907
277 ÷ 189 = 1.46560846561
406 ÷ 277 = 1.46570397112
595 ÷ 406 = 1.46551724138
872 ÷ 595 = 1.46554621849
1278 ÷ 872 = 1.46559633028
1873 ÷ 1278 = 1.46557120501
2745 ÷ 1873 = 1.46556326749
4023 ÷ 2745 = 1.46557377049
5896 ÷ 4023 = 1.46557295551
8641 ÷ 5896 = 1.46556987788
12664 ÷ 8641 = 1.46557111445
18560 ÷ 12664 = 1.46557169931
27201 ÷ 18560 = 1.46557112069
39865 ÷ 27201 = 1.46557111871
58425 ÷ 39865 = 1.46557130315
85626 ÷ 58425 = 1.46557124519
125491 ÷ 85626 = 1.46557120501
183916 ÷ 125491 = 1.46557123618
269542 ÷ 183916 = 1.46557123904
395033 ÷ 269542 = 1.46557122823
578949 ÷ 395033 = 1.46557123076
848491 ÷ 578949 = 1.46557123339
1243524 ÷ 848491 = 1.46557123175
1822473 ÷ 1243524 = 1.46557123144
2670964 ÷ 1822473 = 1.46557123206
3914488 ÷ 2670964 = 1.46557123196
5736961 ÷ 3914488 = 1.46557123179
8407925 ÷ 5736961 = 1.46557123188
12322413 ÷ 8407925 = 1.46557123190
18059374 ÷ 12322413 = 1.46557123187
26467299 ÷ 18059374 = 1.46557123187
38789712 ÷ 26467299 = 1.46557123188
56849086 ÷ 38789712 = 1.46557123188
83316385 ÷ 56849086 = 1.46557123188
122106097 ÷ 83316385 = 1.46557123188
178955183 ÷ 122106097 = 1.46557123188
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):
- Bernoulli's Method for Single Dominant Zero (p. 80)