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It is rather well-known that,

$e^{\pi\sqrt{43}} \approx 960^3 + 743.999\ldots$

$e^{\pi\sqrt{67}} \approx 5280^3 + 743.99999\ldots$

$e^{\pi\sqrt{163}} \approx 640320^3 + 743.999999999999\ldots$

Not so well-known is,

$e^{\pi\sqrt{43}} \approx (5x_1)^3 + 6.000000010\ldots$

$e^{\pi\sqrt{67}} \approx (5x_2)^3 + 6.000000000061\ldots$

$e^{\pi\sqrt{163}} \approx (5x_3)^3 + 6.000000000000000034\ldots$

where the $x_i$ is the appropriate root of the sextics,

$5x^6-960x^5-10x^3+1 = 0$

$5x^6-5280x^5-10x^3+1 = 0$

$5x^6-640320x^5-10x^3+1 = 0$

One can see the j-invariants (or at least their cube roots) appearing again. These sextics are solvable in radicals, factoring over $Q(\sqrt{5})$. However, a more interesting field is $Q(\phi)$, with the golden ratio $\phi = (1+\sqrt{5})/2$. Hence, these sextics have the relevant cubic factor,

$5x^3 - 5(53+86\phi)x^2 + 5(\color{blue}{8}+\color{blue}{13}\phi)x - (\color{red}{18}+\color{red}{29}\phi) = 0$

$5x^3 - 20(73+118\phi)x^2 - 20(\color{blue}{21}+\color{blue}{34}\phi)x - (\color{red}{47}+\color{red}{76}\phi) = 0$

$5x^3 - 20(8849+14318\phi)x^2 + 20(\color{blue}{377}+\color{blue}{610}\phi)x - (\color{red}{843}+\color{red}{1364}\phi) = 0$

respectively. Compare the x term with the Fibonacci numbers,

$F_n = 0, 1, 1, 2, 3, 5, \color{blue}{8, 13, 21, 34}, 55, 89, 144, 233, \color{blue}{377, 610},\dots$

and the constant term with the Lucas numbers,

$L_n = 2, 1, 3, 4, 7, 11, \color{red}{18, 29, 47, 76}, 123, 199, 322, 521, \color{red}{843, 1364},\dots$

Why, oh, why?

P.S. These can be easily verified in Mathematica using the Resultant[] function,

Resultant[$5x^3 - 5(53+86\phi)x^2 + 5(8+13\phi)x - (18+29\phi)$, $\phi^2-\phi-1$, $\phi$]

which eliminates $\phi$ and restores the original sextic. (Similarly for the other two.)

  • 18
    Would it be slightly less mysterious if I remarked that $F_{n-1} + F_n \phi = \phi^n$ for $n \ge 1$ (and presumably a similar identity holds for the Lucas numbers)?2011-05-22
  • 10
    I don't get the bit about ${\bf Q}(\phi)$ being a more interesting field than ${\bf Q}(\sqrt5)$ --- it's the **same** field, innit?2011-05-23
  • 2
    1 To Yuan: Oh, I should have seen that! Thus, $5x^3 - 5(53+86\phi)x^2 + 5\phi^7x - \sqrt{5}\phi^7 = 0$, $5x^3 - 20(73+118\phi)x^2 - 20\phi^9x - \sqrt{5}\phi^9 = 0$, $5x^3 - 20(8849+14318φ)x^2 + 20\phi^{15}x - \sqrt{5}\phi^{15} = 0$. 2. To Gerry: My apologies. You are right. My intent was that if one factors it over $Q(\phi)$ vs $Q(\sqrt{5})$, the polynomial $F_{n-1}+F_n\phi$ is more recognizable than some polynomial $a+b\sqrt{5}$.2011-05-23
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    @QiaochuYuan Check [this out](http://math.stackexchange.com/questions/106049/another-way-to-go-about-proving-the-limit-of-fibonaccis-sequence-quotient).2012-04-05
  • 0
    The Fibonacci number $377$ also appears in Ptolemy's approximation to $\pi$, which can be _explained_ by this series of small positive terms $$\sum_{k=1}^\infty \frac{5040}{(4 k+2) (4 k+3) (4 k+4) (4 k+5) (4 k+7) (4 k+8) (4 k+9) (4 k+10)} = \frac{377}{120}-\pi$$2016-03-18

2 Answers 2