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To make some interesting recursive function, I generalized Fibonacci numbers to a function $f(x)$ such that satisfies the following condition:

Given a function $g(x)$, such that $g(0)=0$ and $g(1)=1$, defined on the interval $x\in [0,2)$, $f(x+2)=f(x+1)+f(x)$ for $\forall x\in \mathbb{R}$, where $f(x)=g(x)$ on the interval $x\in [0,2)$

I believe that this function has a boring mechanism and there is not so interesting feature in it. Do you think the function is worth investigating deeply?

Math.StackExchange
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    Consider $$f(x)=\frac{\varphi^x-\cos\pi x\cdot \varphi^{-x}}{\sqrt 5}$$ – Pedro Jul 12 '13 at 22:27
  • Going from a discrete case to continuous case is usually interesting and non-trivial. So perhaps finding a Fibonacci function that somehow becomes a `natural' extension of Fibonacci numbers will be a nice question. What choice of $g$ is the natural choice? What extension of recurrence relation is interesting or natural? – Maesumi Jul 13 '13 at 00:29

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You did not define a function, but rather a family of functions. For each $g$ you get a different $f$. Functionally, there is nothing going on here. For each $g(x)$ and $(x+1)$ given by $g$, you generate recursively a Fibonacci-like sequence. So what is going here is that you have a whole family of such sequences, given by the initial values specified by $g$. Since $g(0)=0$ and $g(1)=1$, one of those sequences is actually the Fibonacci sequence. The others may or may not be.

Ittay Weiss
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