# Properties of the super-golden ratio?

1. Jul 16, 2011

### dimension10

The Supergolden ratio is the solution of x3=x2+1.

$$\psi = \left({{\sqrt{31}}\over{2\ \times 3^{{{3}\over{2}}}}} {{29}\over{54}} \right)^{{{1}\over{3}}} {{1}\over{9\,\left({{\sqrt{31}}\over{2\ \times3^{ {{3}\over{2}}}}} {{29}\over{54}}\right)^{{{1}\over{3}}}}} {{1}\over{ 3}}\approx 1.46557123187675$$

Can anyone tell me some of its properties, Thanks.

2. Aug 20, 2011

### SubZir0

Hmm, like the golden rectangle, the super golden rectangle has square related recursive properties:

-> Say you have the supergolden rect and you draw a line in it to make a square, then you dot a line from the corner of the rect/square to the opposite corner of the rect you will have an intersecting point. drawing a line accross the not square part of your rectangle and you're left with a tall rect and the supergolden rect (a size down).

sorry i'm not better at explaining things, it's related to the cattle sequence and can't be made using a compass like the golden rect.

3. Aug 21, 2011

### RamaWolf

Could be, your formula is wrong ?
Anyway, the value is OK

See more at www.wolframalpha.com and enter:

Solve[x^3 == x^2 + 1, x]

4. Aug 23, 2011

### RamaWolf

$\psi = \frac{1}{6}*(2 + (116-12*\sqrt{93})^{\frac{1}{3}}+ (116+12*\sqrt{93})^{\frac{1}{3}})$ ≈1.46557123187677

calculated via PB EXT arithmetic as:

e(1) = 116 - 12*SQR(93)
e(2) = 116 + 12*SQR(93)
result = (2+e(1)^(1/3)+e(2)^(1/3))/6.0

5. Aug 24, 2011

### RamaWolf

Perhaps I should note, that the equation has two additional (imaginary) solutions:

$\frac{1}{12}*(4-\sqrt[3]{116-12*\sqrt{93}}-\sqrt[3]{116+12*\sqrt{93}}\pm \sqrt{3}*(\sqrt[3]{116-12*\sqrt{93}}-\sqrt[3]{116+12*\sqrt{93}})*I)$

The numerical approximative values are:

-0.232785615938384 $\pm$ 0.792551992515448 * I

6. Aug 25, 2011

### dimension10

I think I made a mistake. It should be

$$x= \left({{\sqrt{31}}\over{2 \times 3^{{{3}\over{2}}}}} {{29}\over{54}} \right)^{{{1}\over{3}}} + {{1}\over{9\,\left({{\sqrt{31}}\over{2\times 3^{ {{3}\over{2}}}}} {{29}\over{54}}\right)^{{{1}\over{3}}}}}+ {{1}\over{ 3}}$$