Fibonacci numbers extended I am so excited and enjoyed the both the proofs of my previous question on Fibonacci series. I am so interested and fascinating person on fib series/functions. I use to do some rough work in my leisure time on the fib series. I encountered these new two following problems. As per my rough work and my calculator, I am very much correct to state the following questions, whereas mathematically, I am again failed to produce a proof of below. Can you help me please…


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*Let $f(x) = M(x) g(x)$ is a function with $M(x)$ is an f-even function and $g(x)$ is some continuous function. The product of f-even and continuous function is Fibonacci function if and only if continuous function should be a Fibonacci function.  How to justify the above cited statement mathematically. Can we see such facts if we change f-even function to f-odd function of $M(x)$. If yes, what will happen?

*Can we define the limit of the quotient of  $f(x+1)/f(x)$, where $x$ tends to infinity? And the limit of this quotient is close to $1.6$, being the $f$ is Fibonacci sequence.
Thank you so much for all members.
Fibonacci function I defined as per the request from mathematicians.
Please look the definition of Fibonacci function at bellow my one of the comment and kindly answer my first question of this post.
Fibonacci function means, for all x in R, if f: R to R and satisfies f(x+2) = f(x+1) + f(x). Now, let me take f(x) = $M^x$  for some M > 0 As per my definition, f(x+2) = $M^x$  $M^2$  and f(x +1) = $M^x$  M and f(x) = $M^x$ Now, by Fibonacci function definition, we see; $M^2$  = M + 1 Which is same as (1 + square root 5)/2. Now our f(x) = $M^x$  = [(1 + square root of 5)/2]^x. I think, now one can answer my first question
Thank you.
 A: There are various ways to extend the notion of Fibonacci number. One of the more common ones is to let 
$$F(x)=\frac{\phi^x-\cos(\pi x)\phi^{-x}}{\sqrt{5}},$$
where $\phi=\frac{1+\sqrt{5}}{2}$.  This gets over the difficulty that $\left(\frac{1-\sqrt{5}}{2}\right)^x$ makes no sense for most real values of $x$.
Indeed one can in a similar way define $F(z)$ for complex $z$. 
The required limit result is an almost immediate consequence of the definition. 
A: I don't know what "f-even" means, but it is well-known that $$\lim_{x\to\infty}{f(x+1)\over f(x)} = {1 + \sqrt 5 \over 2 } \approx 1.618\ldots$$
Wikipedia has a number of proofs of this and discussion of related matters.  Here's a simple argument:
$$\begin{eqnarray}
L & = & \lim_{x\to\infty}{f(x+1)\over f(x)} \\
&  = & \lim_{x\to\infty}{f(x)+f(x-1)\over f(x)} \\
& = & \lim_{x\to\infty}1 + {f(x-1)\over f(x)} \\
& = & 1 + \lim_{x\to\infty}{f(x-1)\over f(x)} \\
& = & 1 + \lim_{x\to\infty}{f(x)\over f(x+1)} \\
& = & 1 + \lim_{x\to\infty}\left({f(x+1)\over f(x)}\right)^{-1} \\
& = & 1 + \left(\lim_{x\to\infty}{f(x+1)\over f(x)}\right)^{-1} \\
& = & 1 + L^{-1}
\end{eqnarray}$$
Solving for $L$ in $L=1 + L^{-1}$ gives:
$$L={1 \pm \sqrt 5\over 2}$$
and we discard the superfluous negative solution $\frac12(1-\sqrt 5)$.
