Super Golden Ratio

[Thomas Klemm]

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 }
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)