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Properties of the super-golden ratio?

  1. Jul 16, 2011 #1
    The Supergolden ratio is the solution of x3=x2+1.

    [tex]\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[/tex]

    Can anyone tell me some of its properties, Thanks.
  2. jcsd
  3. Aug 20, 2011 #2
    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.
  4. Aug 21, 2011 #3
    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]
  5. Aug 23, 2011 #4
    For your convenience:

    [itex]\psi = \frac{1}{6}*(2 + (116-12*\sqrt{93})^{\frac{1}{3}}+ (116+12*\sqrt{93})^{\frac{1}{3}})[/itex] ≈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
  6. Aug 24, 2011 #5
    Perhaps I should note, that the equation has two additional (imaginary) solutions:

    [itex]\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)[/itex]

    The numerical approximative values are:

    -0.232785615938384 [itex]\pm[/itex] 0.792551992515448 * I
  7. Aug 25, 2011 #6
    I think I made a mistake. It should be

    [tex]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}} [/tex]
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