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As I've said before in other questions I've posed here, I'm an armchair mathematician and don't fully grasp a lot of more advanced concepts (I cap out around early AB Calc), and a lot of my knowledge/interest in maths comes from Numberphile, Mathologer, and other channels like that.

Anyway, I found a formula for the Fibonacci sequence: $$F(x)=\frac{\phi^x-\phi^{-x}\cos(x\pi)}{\sqrt{5}}$$

Out of curiosity, I wanted to find its inverse so that I can find values of the Fibonacci sequence that equate certain numbers (viz, $\pi$, $\tau$, $\phi$, &c).

To do this, I changed $F(x)$ to $y$ and then just flipped the variables, giving: $$x=\frac{\phi^y-\phi^{-y}\cos(y\pi)}{\sqrt{5}}$$

Obviously, the first step is to multiply both sides of the equation by $\sqrt{5}$ yielding: $$x\sqrt{5}=\phi^y-\phi^{-y}\cos(y\pi)$$

It's here that I'm stuck. My instinct tells me to take the $\ln$ of both sides to get the $y$'s to drop down yielding (I think): $$\ln\left(x\sqrt{5}\right)=y\ln(\phi)+y\ln(\phi)\cos(y\pi)$$

but I'm also concerned about the $\cos(y\pi)$ because (a) I don't know if that would be included in the $\ln$ function and (b) because I don't know how to get rid of it and if that would complicate getting the $y$'s out of the exponents.

Is there even a way to solve for $y$, or is it condemned to be in terms of $x$?

Тyma Gaidash
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Curtis Sheppard
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  • Why do you need the $\cos(x\pi)$ part? It is always 1 for integer $x$. – Joshua Wang Jun 16 '21 at 01:43
  • Also, $\ln(a+b)\neq \ln(a) + \ln(b)$ – Joshua Wang Jun 16 '21 at 01:44
  • The $cos(x\pi)$ is used to flip the sign between positive and negative. $(\mod(x,2)=0\Rightarrow\cos(x\pi)=1)\land)\mod(x,2)=1\Rightarrow\cos(x\pi)=-1)$ – Curtis Sheppard Jun 16 '21 at 01:49
  • Regarding $\ln(a+b)\neq\ln(a)+\ln(b)$: I think that perfectly demonstrates what I mean about being an armchair mathematician with a knowledge cap XD

    The suggestions I've gotten in the past have indicated that I should at least try to solve the problem myself before asking the question and share the work that I've done. That's what I came up with, hence the need for help ^_^'

     – Curtis Sheppard Jun 16 '21 at 01:51
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    Let $z=\phi^y.$ Then $z^2-z(x\sqrt 5)-\cos \pi x=0,$ which is a quadratic equation in $z.$ Find the positive solution $z$ (since $z=\phi^y> 0$) using the Quadratic Formula. Then $\log z=y\log \phi=\log z$, so $y=(\log z)/\log\phi.$ – DanielWainfleet Jun 16 '21 at 05:13
  • @DanielWainfleet But isn't it $\cos \pi y$ rather than $\cos \pi x$? – Etemon Jun 16 '21 at 09:11
  • @Soheil . Not in the title nor the first displayed line . – DanielWainfleet Jun 18 '21 at 04:26
  • Should this question a duplicate of this one? – Тyma Gaidash Aug 29 '23 at 23:44

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