Continuous Fibonacci number F(n) How is a continuous Fibonacci number $F_n$ defined? (like the Gamma function for factorials)
Tried with Binet's formula (Wiki) , but to no luck.
F[n_] = N[GoldenRatio^n - 1/GoldenRatio^n]/Sqrt[5]
Plot[F[n], {n, 3., 12.}, GridLines -> Automatic]
FibonRatio[n_] = N[Fibonacci[n + 1]/Fibonacci[n]];
Plot[{FibonRatio[n], N[GoldenRatio], 1.7}, {n, 3., 12.}, 
 GridLines -> Automatic]

Also using this continuous function definition how is it proved that
$$\lim _{n\rightarrow \infty}\dfrac{F_{n+1}}{F_n} = \phi$$
( where $\phi$ is the GoldenRatio) ?
Appreciate your comments.
 A: $F_n=\frac{\phi^n-\cos(\pi n)\phi^{-n}}{\sqrt{5}},$ with $\phi$ being the golden ratio. Here $n$ can be also complex.
You can also rewrite the ratio as
$$\frac{F_{n+1}}{F_n}=\phi\left(\frac{1+(-1)^{n+1}\phi^{-2(n+1)}}{1+(-1)^{n+1}\phi^{-2n}}\right),$$
where it easier to show that the ratio converges to $\phi$ and maybe you like it for calculations. But for the proof only I would rather prefer the other answers to be honest.
A: Since you are using Mathematica, judging by the code in the OP, you can easily obtain the actual expression used by Mathematica for evaluating Fibonacci numbers of non-integer index:
FunctionExpand[Fibonacci[n]]
   ((1/2 (1 + Sqrt[5]))^n - (2/(1 + Sqrt[5]))^n Cos[n π])/Sqrt[5]

In more conventional notation ($\phi$ is the golden ratio, as usual):
$$F_n=\frac{\phi ^n-\phi ^{-n} \cos (\pi  n)}{\sqrt{5}}$$
(This is also the expression given in the Wolfram Functions site and CostaZach's answer.)
This of course doesn't imply that this is the only way to define the Fibonacci numbers for non-integer index. For instance, this function uses
$$f_n=\frac{(1 + \sqrt{5})^n - (1 - \sqrt{5})^n}{2^n\sqrt{5}}$$
but unlike the previous formula, this becomes complex-valued in general for real non-integer argument.
A: $\DeclareMathOperator{\bbN}{\mathbb{N}}$$\newcommand{\digsum}{\text{digitsum}} \DeclareMathOperator{\bbO}{\mathbb{O}} \DeclareMathOperator{\bbE}{\mathbb{E}}$$\blacksquare$ Notation: Let us denote:

*

*$\bbO := $ the set of all odd natural numbe\rs

*$\bbE := $ the set of all even natural numbers


Hint: Let $F_0 = 0$ and $F_1 = 1$. We can see the following:
$$ F_{n + 1} = F_{n} + F_{n - 1} \quad \text{for } n \geqslant 1  $$
$$ \implies \frac{F_{n + 1}}{F_n} = 1 + \frac{F_{n-1}}{F_n} = 1 + \frac{1}{\frac{F_{n}}{F_{n - 1}}} = 1 + \frac{1}{1 + \frac{F_{n-2}}{F_{n - 1}}} = \cdots = 1 + 
\frac{1}{1 + \frac{1}{ 1 + \frac{1}{1 + \frac{1}{1 + \cdots (n\text{ times})}}}} $$

Lemma 1: The sequence $\{ \xi_n\}_{n=1}^\infty := \left\{ \dfrac{F_{n + 1}}{F_n} \right\}_{n = 1}^\infty$ has the following properties:

*

*The even subsequence $\{ \xi_{2k}\}_{k=1}^\infty := \left\{ \dfrac{F_{2k + 1}}{F_{2k}} \right\}_{k = 1}^\infty$ increases.


*The odd subsequence $\{ \xi_{2k + 1}\}_{k=0}^\infty := \left\{ \dfrac{F_{2k + 2}}{F_{2k + 1}} \right\}_{k = 0}^\infty$ decreases.
Proof: The proof will be immediate from the relation $~F_{n}^2 + (-1)^n = F_{n + 1}F_{n - 1}  ~$ for any $n \geqslant 1$ [prove it using induction].
Now we have that $$ \xi_{n+1} - \xi_n = \frac{F_{n + 1}}{F_n} - \frac{F_n}{F_{n - 1}} = \frac{F_{n +1}F_{n - 1} - F_n^2 }{F_n F_{n - 1}} = \frac{(-1)^n}{F_nF_{n - 1}}  \begin{cases} >0 & \text{if } n \in \bbE\\ < 0 & \text{if } n \in \bbO  \end{cases}  $$
Thus we proved our claim. $\hspace{14cm} \blacksquare$

Lemma 2: The sequence $\{ \xi_n\}_{n = 1}^\infty$ is bounded.
Proof: We know that $\{ F_n\}_{n = 0}^\infty$ is an increasing sequence. Note that $$ F_{n + 1} = F_{n} +F_{n - 1} \implies \xi_n = 1 + \frac{F_{n - 1}}{F_n} < 1 + 1 = 2 $$
And trivially $ 0 < \xi_n < 2 ~$ for all $n \in \bbN$
Thus we have that $$0 <\frac{F_{n+1}}{F_{n}} <  {2}  $$
Hence it's bounded. $\hspace{15cm} \blacksquare$

Lemma 3: The following holds:
$$ \lim_{n \to \infty} \frac{F_{n + 1}}{F_n} ~\text{  exists } $$
Proof: Follows from Lemma 1 and Lemma 2 $\hspace{7cm} \blacksquare$
So, let's observe that $$\lim\limits_{n \to \infty} \frac{F_{n + 1}}{F_n } = L  = 1 + 
\frac{1}{1 + \frac{1}{ 1 + \frac{1}{1 + \frac{1}{1 + \cdots (\infty\text{ times})}}}} = \frac{1}{1 + L} $$
$$ \implies L^2 - L - 1 = 0 \implies L = \frac{1 + \sqrt{5}}{2} = \phi $$
A: I think the reason you aren't able to get a continuous graph when plotting the binet is because the $\left(\frac{1-\sqrt{5}}{2}\right)^n$ term is unreal for most noninteger values of $n$.
I don't think there exists any continuous formula for the Fibonacci Sequence, but $\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{x}+\left(\frac{\sqrt{5}-1}{2}\right)^{x}\right)$ and $\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{x}-\left(\frac{\sqrt{5}-1}{2}\right)^{x}\right)$ are both good continuous approximations and are exact for odd and even terms respectively.
A less rigorous approach for proving the limit of the ratio of subsequent terms is to note that for $x>>1$ $F_n\approx \frac{\phi^n}{\sqrt{5}}$. So as $x\to\infty$, $\frac{F_{n+1}}{F_n}\to \frac{\phi^{n+1}}{\phi^n}=\phi$
A: Suppose we have a sequence $\{ a_n \}_{n \geq 1}$ obeying a linear recurrence of the form $$a_n = p a_{n-1} + q a_{n-2}$$ for some constants $p, q$. Then if the characteristic equation $$\lambda^2 = p \lambda + q$$ has distinct roots, then every solution to the recurrence is of the form $$a_n = c_1 \lambda_1^n + c_2 \lambda_2^n,$$ where $\lambda_1, \lambda_2$ are the two roots of the characteristic equation, and $c_1, c_2$ are constants chosen to match initial conditions on the sequence (usually, specified values of $a_1$ and $a_2$).
In the case of the Fibonacci sequence, the recurrence is $a_n = a_{n-1} + a_{n-2}$, with initial conditions $a_1 = a_2 = 1$. The characteristic equation is $\lambda^2 = \lambda + 1$, with roots $\lambda = \frac{1 \pm \sqrt{5}}{2}$.
Writing $\lambda_1 = \phi = \frac{1 + \sqrt{5}}{2} \approx 1.618...$, the other root is $\lambda_2 = -1/\phi = \frac{1 - \sqrt{5}}{2} \approx -0.618...$, and the constants $c_1, c_2$ making $a_1 = a_2 = 1$ are $c_1 = 1/\sqrt{5}, c_2 = -1/\sqrt{5}$. From this we find the formula, valid for all $n \geq 1$, $$a_n = F_n = \frac{\phi^n - (-\phi)^{-n}}{\sqrt{5}},$$ and one desired continuous extension is clearly the real part $$F_x = \Re(\frac{\phi^x - (-\phi)^{-x}}{\sqrt{5}}).$$
For large values of $x$, the $\phi^{-x}$ term (which is also the source of the imaginary part) tends to zero, so $F_x \approx \phi^x/\sqrt{5}$, from which we easily derive $$\lim_{x \to \infty} \frac{F_{x+1}}{F_x} = \phi.$$
