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I was wondering if there was a relation between a Fibonacci number and its position. Is there a function $f(n)$ such that $$f(n)=F_n$$ where $F_n$ is the nth Fibonacci number?

PunkZebra
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  • do you know $F_{n}$ closed formula ? – Khosrotash Jun 15 '15 at 10:45
  • You mean $F_n=F_{n-1}+F_{n-2}$? – PunkZebra Jun 15 '15 at 10:48
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    no , I mean $$f_{n}=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^n+(\frac{1-\sqrt{5}}{2})^n)$$ – Khosrotash Jun 15 '15 at 10:52
  • Note that, even if there wasn't a closed form, $f$ where $f(n)=F_n$ is still a function. A function is any rule that turns a number into a number (along with a domain). So $f(n)$, the way you've described it, is a function. (You seem to have the misconception that a function is any combination of mathematical symbols… this is generally called a closed form formula.) – Akiva Weinberger Jun 15 '15 at 19:33

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There is also

$$\begin{bmatrix} F(n + 1) & F(n) \\ F(n) & F(n - 1) \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n$$

DanielV
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That f(n) you are looking for is the closed-form expression, you can verify its definition here:

$$F_n = \frac{\varphi^n - (1-\varphi)^n}{\sqrt{5}}, \quad\text{where}\quad \varphi = \frac{1+\sqrt{5}}{2}.$$

Daniel Fischer
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iadvd
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