4
$\begingroup$

We denote the $n$th term of Fibonacci number with $F_n$. Assume that $\alpha=\frac{1+\sqrt{5}}{2}$. With simulation, I found the following relation between Fibonacci number and the golden section $$ \mid \frac{F_{n+1}}{F_{n}}-\alpha\mid \, \approx \, \frac{1}{(F_n)^2}~. $$ Is there a analytical method that we can proof the mentioned formula. I would greatly appreciate for any suggestions.

Maple simulation

Edit: First I want to gratitude from Milo Brandt for nice answer. In continue, i want to generalize my question.

One of the most important generalization of the classical Fibonacci numbers is the Fibonacci $p$-step numbers that is defined as follows $$ \begin{equation}\label{cp26} F_n^{(p)}=F_{n-1}^{(p)}+F_{n-2}^{(p)}+\cdots+F_{n-p}^{(p)}\, . \end{equation} $$ With boundary conditions $$ F_{0}^{(p)}=0\quad , \quad F_{1}^{(p)}=0\quad ,\, \cdots\, ,\quad F_{p-2}^{(p)}=0\quad , \quad F_{p-1}^{(p)}=1\, . $$

We can get the limit value of Fibonacci $p$-step numbers by inverse of solution of equation $x^{p+1}-2\, x+1=0$ in the interval $(0,1)$. We denote the limit value of Fibonacci $p$-step numbers with $\alpha_p$. In fact, $\alpha_p$ is defined in the following form $$ \alpha_p=\displaystyle{\lim_{n\rightarrow\infty}}\quad \frac{F^{(p)}_{n+1}}{F^{(p)}_{n}}~. $$ The generalization of the above formula is $$ \mid \frac{F^{(p)}_{n+1}}{F^{(p)}_{n}}-\alpha_p\mid \, \approx \, {F^{(p)}_{n}}^{-{\displaystyle{(\frac{p}{p-1})}}}~. $$

For example for the case $p=4$, we have

Fibonacci $4$-step numbers

$\endgroup$
4
  • $\begingroup$ You get this because the Fibonacci series tends to the golden ratio, so subtracting alpha tends to zero, and so does the reciprocal function. There's nothing deep about it. $\endgroup$ – Nij Jul 23 '16 at 22:08
  • $\begingroup$ My question is a especial case of the following question math.stackexchange.com/questions/1837391/… $\endgroup$ – Amin235 Jul 23 '16 at 22:13
  • 1
    $\begingroup$ @Amin23 It'd be better to have the generalization as a separate post - its answer isn't particularly related to the answer to your original question. Unless I'm terribly mistaken, your conjecture is true for $p=3$ and false for $p=4$. (It seems to be more of a coincidence than anything else, but I'm not certain) $\endgroup$ – Milo Brandt Jul 23 '16 at 23:02
  • $\begingroup$ @MiloBrandt This conjecture is true for all case of $p$. Please see image file for the case $p=4$. In fact, this conjecture is an especial case of the following question math.stackexchange.com/questions/1837391/… $\endgroup$ – Amin235 Jul 24 '16 at 5:15
10
$\begingroup$

The most direct way to deal with the Fibonacci numbers is to use Binet's formula: $$F_n=\frac{\varphi^n - (-\varphi)^{-n}}{\sqrt{5}}$$ where $\varphi=\frac{1+\sqrt{5}}2$. So, the ratio of $\frac{F_{n+1}}{F_n}$ can be written in closed form as: $$\frac{F_{n+1}}{F_n}=\frac{\varphi^{n+1} - (-\varphi)^{-n-1}}{\varphi^n - (-\varphi)^{-n}}$$ Note that this obviously tends towards $\varphi$ as $n$ goes to $\infty$, since the $(-\varphi)^{-n}$ terms quickly go to zero. Now, you additionally want to show that the difference between this ratio and $\varphi$ shrinks with $\frac{1}{F_n^2}$. To do this, let us take that difference symbolically: $$\frac{F_{n+1}}{F_n}-\varphi = \frac{\varphi^{n+1}-(-\varphi)^{-n-1}}{\varphi^n-(-\varphi)^{-n}}-\frac{\varphi^{n+1}+(-\varphi)^{-n+1}}{\varphi^n-(-\varphi)^{-n}}=\frac{-(-\varphi)^{-n+1}-(-\varphi)^{-n-1}}{\varphi^n-(-\varphi)^{-n}}$$ Then, we use that $(\varphi)^{-1}+\varphi=\sqrt{5}$ to simplify to $$\frac{F_{n+1}}{F_n}-\varphi = \frac{\sqrt{5}(-\varphi)^{-n}}{\varphi^n - (-\varphi)^{-n}}\approx \frac{\sqrt{5}(-1)^n}{\varphi^{2n}}\approx \frac{(-1)^n}{\sqrt{5}F_n^2}$$ where the $\approx$ signs indicate that the ratio of the two sides of the "equation" tends to $1$ as $n$ goes to $\infty$. We use that $F_n\approx \frac{\varphi^n}{\sqrt{5}}$, which implies $F_n^2\approx \frac{\varphi^{2n}}{5}$

$\endgroup$
2
  • 1
    $\begingroup$ Actually, you must have made a mistake because the last exact equality doesn't hold. Also, assuming you're dealing with asymptotic equivalence (since trivially the expressions in the OP are both $\approx0$), one actually has $$\frac{F_{n+1}}{F_n}-\varphi\sim\frac{(-1)^n}{\sqrt{5} F_n^2}.$$ $\endgroup$ – Vincenzo Oliva Jul 24 '16 at 9:00
  • 1
    $\begingroup$ @VincenzoOliva Thank you - I found the error and fixed it. $\endgroup$ – Milo Brandt Jul 24 '16 at 14:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.