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I was interested by another user's question on finding such an inverse and in particular noted Will Orrick's comment in the best answer that one can square both sides to obtain a quartic.I thought I'd solve that quartic. For clarity, here is the work in full; the customary notation $F_n$ has been replaced with $F$ for simplicity. $\phi$ is the golden ratio $\frac{1 + \sqrt{5}}{2}$, $n$ is the position in the sequence, and $F$ is the $n$th Fibonacci number.

$$ \begin{align} n &= \log_\phi\frac{F\sqrt{5} + \sqrt{5F^2 + 4(-1)^n}}{2}\\ \phi^n &= \frac{F\sqrt{5} + \sqrt{5F^2 + 4(-1)^n}}{2}\\ 2\phi^n &= F\sqrt{5} + \sqrt{5F^2 + 4(-1)^n}\\ 2\phi^n - F\sqrt{5} &= \sqrt{5F^2 + 4(-1)^n}\\ 4\phi^{2n} - 4\phi^nF\sqrt{5} + 5F^2 &= 5F^2 + 4(-1)^n\\ 4\phi^{2n} - 4\phi^nF\sqrt{5} &= 4(-1)^n\\ \phi^{2n} - \phi^nF\sqrt{5} &= (-1)^n\\ \phi^{4n} - 2\phi^{3n}F\sqrt{5} + 5\phi^{2n}F^2 &= 1\\ \phi^{4n} - 2\phi^{3n}F\sqrt{5} + 5\phi^{2n}F^2 - 1 &= 0 \end{align} $$

At this point, we can apply the quartic formula (edit: for clarity I am showing all the steps of applying the formula):

$$a = 1, b = 2F\sqrt{5}, c = 5F^2, d = 0, e = -1\\ \begin{align} u &= \frac{3(2F\sqrt{5})^2 - 8(1)(5F^2)}{12(1^2)} = \frac{60F^2 - 40F^2}{12} = {5F^2}{3}\\ \Delta_0 &= (5F^2)^2 - 3(2F\sqrt{5})(0) + 12(1)(-1) = 25F^4 - 12\\ \Delta_1 &= 2(5F^2)^3 - 9(2F\sqrt{5})(5F^2)(0) + 27(2F\sqrt{5})^2(-1) - 72(5F^2)(-1) = 250F^6 - 180F^2\\ Q &= \sqrt[3]{\frac{250F^6 - 180F^2 + \sqrt{(250F^6 - 180F^2)^2 - 4(25F^4 - 12)^3}}{2}} = \sqrt[3]{125F^6 - 90F^2 + 6\sqrt{48 - 75F^4}}\\ v &= \frac{Q^2 + \Delta_0}{3Q}\\ w &= 0 \end{align} $$

$$n_{1,2} = \log_\phi \frac{1}{2}\left(F\sqrt{5} - \sqrt{\frac{5F^2}{3} + \frac{\left(\sqrt[3]{125F^6 - 90F^2 + 6\sqrt{48 - 75F^4}}\right)^2 + 25F^4 - 12}{3\sqrt[3]{125F^6 - 90F^2 + 6\sqrt{48 - 75F^4}}}}\\ \pm \sqrt{\frac{10F^2}{3} - \frac{\left(\sqrt[3]{125F^6 - 90F^2 + 6\sqrt{48 - 75F^4}}\right)^2 + 25F^4 - 12}{3\sqrt[3]{125F^6 - 90F^2 + 6\sqrt{48 - 75F^4}}}}\right)$$

The third and fourth solutions are given by the first expression, but with a plus sign after $F\sqrt{5}$.

Are my calculations correct (they were done entirely by hand), is this a correct inverse to the Binet formula, and if so, how do the multiple solutions correspond to the actual solutions of n for F?

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Lee Sleek
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  • I think separate formulas for odd and even $n$ would be only quadratic. Would these be of interest to you? – coffeemath Jun 11 '13 at 05:46
  • @coffeemath That's true, but Linus (who asked the previous question) was adamant in wanting something more general. – Lee Sleek Jun 11 '13 at 14:30
  • On your step in the first part from line 7 to line 8, the $(-1)^n$ gets squared. The effect of this is that no amount of manipulation after that can distinguish between $n$ being odd or even, so it seems there will necessarily be an ambiguity in the formula that results, where sometimes one sign choice is needed and sometimes the other. – coffeemath Jun 13 '13 at 23:39
  • The point of this question is to discover what the sign change means. – Lee Sleek Jun 13 '13 at 23:57
  • By "what the sign change means" do you mean "how can the specific choice of sign be computed from $F$ alone"? Otherwise I don't know the sense of the phrase in quotes... – coffeemath Jun 14 '13 at 01:21
  • I agree that was poor wording. Sorry about that. I asked this question because I'd like to know if this is a correct formula at all, if my calculations were correct, and if this is a correct inverse formula then how the four solutions correspond to the actual position in the sequence of a Fibonacci number. (This question has received hardly any attention from anyone but us and I'm considering starting a bounty on it.) – Lee Sleek Jun 14 '13 at 04:36
  • The proof is in the pudding. By that I mean, try your formula(s) on various known Fibonacci numbers for $F$, and see whether (when enough digits are taken for the calculation to be accurate, or alternatively when a CAS like maple is used for the simplifications) the correct value of $n$ results. If that happens for even a small number of $F$ choices, I'd be confident about the result. – coffeemath Jun 14 '13 at 05:39

1 Answers1

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I follow all the steps up to line seven of the first part. If we write $x=\phi^n$ then that seventh line reads $$x^2-xF\sqrt{5}=(-1)^n.$$ Then on line eight you "square both sides", the goal being to eliminate the $(-1)^n$ on the right side. But suppose instead of actually distributing the square of the left side, you just keep it algebraically. Then the following line is equivalent to your ninth line (and so must have the same roots, and be the same quartic polynomial): $$ [1] \ \ [x^2 - xF\sqrt{5}]^2-1=0.$$ This is where your first paragraph ends, and you begin applying the solution of the quartic equation. But that is to use a more general tool than necessary for your purpose, since in fact [1] is a difference of squares, so that it factors into two quadratic polynomials. In other words the quartic you have is really a biquadratic, and its solutions are those of the two quadratic factors (two from each, for the total of four). Specifically [1] factors as $$[2]\ \ [x^2-xF\sqrt{5}-1]\cdot [x^2 - x F\sqrt{5}+1]=0.$$ Now depending on whether the original $n$ is odd or even, exactly one of these factors will have one of its roots lead to an integer for $n$, while the other root of that factor will be in a way conjugate to it (but won't lead to an integer $n$). The two quadratic roots of the other factor will not lead to integer $n$ in terms of $F$ either. In sum, only one of the four roots of your quartic will be the desired inverse for $F$, and the quadratically expressed roots of [2] must coincide with the four more convoluted expressions you have obtained using the solution for general quartics.

coffeemath
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  • I was already aware of all the information contained in this answer. Linus, who asked the question that inspired me, was most likely made aware of all of it after accepting the best answer to his question. This does not answer my current question. – Lee Sleek Jun 11 '13 at 14:34
  • @LeeSleek : please have a look at this answer, which is a careful reading of your actual derivation, up to the end of the first paragraph at least. As you'll note, I did not persue the application of the quartic equation since your particular quartic factors into two quadratics, as I note in the (now totally changed) answer above. – coffeemath Jun 14 '13 at 08:18
  • Note: my first comment on this answer was directed at a previous answer, which did not answer the question to my liking. – Lee Sleek Jun 16 '13 at 19:55