Given a value of k, is it possible to express explicitly the collection of x satisfying Fib(x)=k

Question

From Wikipedia, the formula to extend to all real the Fibonacci number is: $$ \text{Fib}(x)=\frac{\varphi^x - \cos(\pi x)\varphi^{-x}}{\sqrt 5} $$ Given a value of k, is it possible to express explicitly the collection of x satisfying Fib(x)=k ?

Answer 1

$\def\Fib{\operatorname{Fib}}$

We apply Lagrange reversion to invert $\Fib(x)$, like Wolfram repository’s inverse Fibonacci. It may be possible to get more branches by substituting $\phi x=\ln(w)+2\pi ik,k\in\Bbb Z$ or solving for $x=\cdots$ and using the complex logarithm, but that is more complicated, so $k=0$:

$$\Fib(\log_\phi(w)))=a\iff w=\sqrt 5a+\frac12w^{-\frac{\pi i}{\ln(\phi)}-1}+\frac12w^{\frac{\pi i }{\ln(\phi)}-1}\implies \Fib^{-1}(x)=\log_\phi\left(\sqrt5x+\sum_{n=1}^\infty\frac1{2^nn!}\left.\frac{d^{n-1}}{dx^{n-1}}\left(x^{-\frac{\pi i}{\ln(\phi)}-1}+x^{\frac{\pi i }{\ln(\phi)}-1}\right)^n \right|_{\sqrt5x}\right)$$

Next use the binomial theorem:

$$\frac{d^{n-1}}{dx^{n-1}}\left(x^{-\frac{\pi i}{\ln(\phi)}-1}+x^{\frac{\pi i }{\ln(\phi)}-1}\right)^n=\sum_{m=0}^n\binom nm \frac{d^{n-1}}{dx^{n-1}}x^{\frac{\pi i (2m-n)}{\ln(\phi)}-n}$$

which makes factorial power $n^{(m)}$ appear. Therefore:

$$\bbox[3px,border: 2px solid lightgreen]{\Fib^{-1}(x)=\log_\phi\left(\sqrt 5x+\sum_{n=1}^\infty\sum_{m=0}^n\left(\frac{\pi i (2m-n)}{\ln(\phi)}-n\right)^{(n-1)}\binom nm\frac{(\sqrt5x)^{\frac{\pi i (2m-n)}{\ln(\phi)}-2n+1}}{2^nn!}\right)}$$

shown here. The result is a double hypergeometric series

Answer 2

In the real domain, if you want to find the zero of function

$$f(x)=\frac{\phi^x - \cos(\pi x)\,\phi^{-x}}{\sqrt 5}-k$$ you can have an explicit expression performing one single iteration of any Newton-like method starting with $$x_0=\frac{\log \left(\sqrt{5} k\right)}{\log (\phi )}$$

Since $$f'(x)=\frac{\pi \phi ^{-x} \sin (\pi x)+\text{csch}^{-1}(2) \phi ^x+\text{csch}^{-1}(2) \phi ^{-x} \cos (\pi x)}{\sqrt{5}}$$ the formula for Newton method is simple but it does not make any problem for Halley or Householder methods.

Trying for a few values of $k$

$$\left( \begin{array}{cccc} k & \text{Newton method} &\text{Halley method} &\text{solution} \\ 2 & 2.99666272790 & 2.99939552636 & 3.00000000000 \\ 3 & 4.00093541961 & 3.99997677784 & 4.00000000000 \\ 4 & 4.54911525254 & 4.54911256276 & 4.54911255651 \\ 5 & 5.00004596343 & 4.99999973617 & 5.00000000000 \\ 6 & 5.39185329558 & 5.39184960448 & 5.39184960691 \\ 7 & 5.72146999558 & 5.72146173169 & 5.72146173497 \\ 8 & 6.00001134392 & 5.99999998654 & 6.00000000000 \\ 9 & 6.24201126307 & 6.24200781881 & 6.24200782135 \\ 10 & 6.45779718802 & 6.45779711744 & 6.45779711745 \\ 11 & 6.65371448777 & 6.65371391099 & 6.65371391090 \\ 12 & 6.83362397630 & 6.83362256091 & 6.83362256070 \\ 13 & 7.00000138968 & 7.00000000012 & 7.00000000000 \\ 14 & 7.15458943766 & 7.15458861266 & 7.15458861262 \\ 15 & 7.29874744351 & 7.29874715848 & 7.29874715847 \\ 16 & 7.43362013074 & 7.43362010332 & 7.43362010332 \\ 17 & 7.56020990738 & 7.56020988947 & 7.56020988947 \\ 18 & 7.67940477808 & 7.67940466250 & 7.67940466251 \\ 19 & 7.79198976833 & 7.79198956259 & 7.79198956261 \\ 20 & 7.89865435506 & 7.89865411580 & 7.89865411582 \\ \end{array} \right)$$