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?
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- $\begingroup$ do you know $F_{n}$ closed formula ? $\endgroup$Khosrotash– Khosrotash2015-06-15 10:45:20 +00:00Commented Jun 15, 2015 at 10:45
- $\begingroup$ You mean $F_n=F_{n-1}+F_{n-2}$? $\endgroup$PunkZebra– PunkZebra2015-06-15 10:48:20 +00:00Commented Jun 15, 2015 at 10:48
- 1$\begingroup$ no , I mean $$f_{n}=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^n+(\frac{1-\sqrt{5}}{2})^n)$$ $\endgroup$Khosrotash– Khosrotash2015-06-15 10:52:34 +00:00Commented Jun 15, 2015 at 10:52
- $\begingroup$ 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.) $\endgroup$Akiva Weinberger– Akiva Weinberger2015-06-15 19:33:46 +00:00Commented Jun 15, 2015 at 19:33
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2 Answers
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There is also
$$\begin{bmatrix} F(n + 1) & F(n) \\ F(n) & F(n - 1) \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n$$
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That f(n) you are looking for is the closed-form expression, you can verify its definition here:
$$F_n = \frac{\varphi^n - (1-\varphi)^n}{\sqrt{5}}, \quad\text{where}\quad \varphi = \frac{1+\sqrt{5}}{2}.$$
