# Finding n in Fibonacci closed loop form

The nth term of the Fibonacci series is given by

$F_{n}$=$\Big\lfloor\frac{\phi^{n}}{\sqrt{5}}+\frac{1}{2}\Big\rfloor$

How do you get the following expression for n from this?

$n=\Big\lfloor\log_{\phi}\Big(F\cdot\sqrt{5}+\frac{1}{2}\Big)\Big\rfloor$

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Do you have some reason to believe that expression for $n$? Did you see it somewhere? –  Gerry Myerson Jun 16 '12 at 12:23
Yes. I got it from the Wikipedia article. en.wikipedia.org/wiki/… –  Guru Prasad Jun 16 '12 at 12:26

Actually you can't get $n=\Big\lfloor\log_{\phi}\Big(F\cdot\sqrt{5}+\frac{1}{2}\Big)\Big\rfloor$ only from $F_n=\Big\lfloor\frac{\phi^{n}}{\sqrt{5}}+\frac{1}{2}\Big\rfloor$. But these two identities can be both deduced from $F_n=\frac{\phi^{n}}{\sqrt{5}}-\frac{\psi^{n}}{\sqrt{5}}$. Here we have $|\psi|<1$, so we can add 1/2 and floor it to clear away the $\psi$ term, which makes the expression nicer in some sense.
From $F_n=\frac{\phi^{n}}{\sqrt{5}}-\frac{\psi^{n}}{\sqrt{5}}$, we get $\sqrt{5}F_n=\phi^{n}-\psi^{n}$. Let's assume $n\geq 2$, then $|\psi^{n}|\leq \psi^{2}<1/2$. (when $n=1,0$, you can directly check the identity which may suit or may not suit)
Thus $\sqrt{5}F_n+\frac{1}{2}>=\phi^{n}$ and trivially $\sqrt{5}F_n+\frac{1}{2}\leq \phi^{n}+1\leq\phi^{n+1}$ since $\phi>1.6$ and $n\geq 2$. Thus $n=\Big\lfloor\log_{\phi}\Big(F\cdot\sqrt{5}+\frac{1}{2}\Big)\Big\rfloor$.
Do you admit $F_n=\frac{\phi^{n}}{\sqrt{5}}-\frac{\psi^{n}}{\sqrt{5}}$? –  caozhu Jun 16 '12 at 12:57