# How does one arrive at a certain expression for the Fibonacci Zeta function?

In this paper by L. Navas, it is described how one can obtain a analytic continuation of the Fibonacci Dirichlet series (though I'm not sure it's actually a Dirichlet Series).

First, the following constants are defined: $\phi = \frac{1 + \sqrt{5}}{2}$, $\phi' = \frac{1-\sqrt{5}}{2}$. Furthermore, we define $f_n$ as the $n$'th Fibonacci number. Then, on pages 411 and 412, the following is derived Binet's formula): $$\begin{split} {f_{n}}^{p} & = \Big{(} \frac{ \phi^{n} - {\phi'}^{n} }{ \sqrt{5}} \Big{)}^{p} = 5^{-\frac{p}{2} } \phi^{np} \Big{(} 1 - \big{(} \frac{\phi'}{\phi} \big{)}^{n} \Big{)}^{p} \qquad (1) \\ & = 5^{- \frac{p}{2} } \phi^{np} \Big{(} 1 - (-1)^{n+1} \frac{1}{{\phi}^{2n}} \Big{)}^{p} \qquad (2). \\ & \end{split}$$

My question is: how does one arrive from equation $(1)$ to equation $(2)$ ?

• Hint: $\;\phi'=-\dfrac 1{\phi}$. – Raymond Manzoni May 14 '15 at 8:09

## 2 Answers

Bearing in mind that $\varphi'=1-\varphi$ and $\frac{1}{\varphi}=\varphi-1$ because $\varphi$ and $\varphi'$ are the roots of the quadratic $X^2-X-1$ one gets

$$\frac{\varphi'}{\varphi}=\frac{1-\varphi}{\varphi}=-\frac{1}{\varphi^2}$$

And so $\left(\frac{\varphi'}{\varphi}\right)^n=(-1)^n\frac{1}{\varphi^{2n}}$ (and you definitely have a typo).

Nothing really deep here (though I think you have a typo): \begin{align} \left( \frac{\varphi'}{\varphi}\right)^n & = \left( \frac{1-\sqrt{5}}{1+\sqrt{5}}\right)^n \\ & = \left( \frac{1-\sqrt{5}}{1+\sqrt{5}}\frac{1+\sqrt{5}}{1+\sqrt{5}}\right)^n \\ & =\left( \frac{-4}{(1+\sqrt{5})^2}\right)^n \\ & = (-1)^n \left( \frac{4}{(1+\sqrt{5})^2} \right)^n \\ & = (-1)^n \frac{1}{\varphi^{2n}} \end{align}