# A pattern appearing in the powers of $\phi$

Alaa Nasr 07/11/2018 at 22:47. 3 answers, 625 views

\begin{align} \phi^5 &= 11,\underline{0}901699\cdots\\ \phi^6 &= 17,\underline{9}44271\cdots\\ \phi^7 &= 29,\underline{6}34441\cdots\\ \phi^8 &= 46,\underline{9}7871\cdots\\ \phi^9 &= 76,\underline{0}1315 \cdots\\ \phi^{10} &= 122,\underline{99}18\cdots\\ \phi^{11} &= 199,\underline{00}502\cdots\\ \phi^{12} &= 321,\underline{99}6894\cdots\\ \phi^{13} &= 521,\underline{00}191\cdots\\ \phi^{14} &= 842,\underline{99}881\cdots\\ \phi^{15} &= 1364,\underline{000}73\cdots\\ \phi^{16} &= 2206,\underline{999}54\cdots\\ \end{align}

Why there is a $0$ $9$ patterns in the powers of the golden ratio

orlp 07/11/2018 at 23:09.

This can be seen from the following formula:

$$L_n = \varphi^n + \frac{1}{(-\varphi)^n}$$

Where $L_n$ are the Lucas numbers, which are integers. Because the term $\dfrac{1}{(-\varphi)^n}$ alternates between a tiny positive and negative value, we see that $\varphi^n$ must be just barely below or above an integer - hence the $.0$ and $.9$ pattern.

lulu 07/11/2018 at 23:19.

$\phi$ is the larger root of $$x^2-x-1=0$$

It's conjugate root is: $$\overline {\phi}=\frac {1-\sqrt 5}2\approx -0.61803$$

From the quadratic, we see that the sequence $$\{a_n\}=\{\phi^n +\bar {\phi}^n\}$$ satisfies the Fibonacci recursion: $$a_{n+1}=a_n+a_{n-1}\quad a_1=1\quad a_2=3$$

Of course $\bar {\phi}^n\to 0$ for large $n$ so we must have that $\phi^n$ is nearly an integer for large $n$. since $\bar {\phi}^n$ alternates sign we see the pattern you have noticed.

RayDansh 07/12/2018 at 09:55.

I cannot provide an explanation for why the $0$ and $9$ patterns appear, but I can give an explanation of why it seems like the powers converge to integers.

If you visit the following website: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/phigits.html

Based on the special properties of $\phi$ (I won't prove it here),

$$\phi^2 -\phi^{-2} = 3,\\\phi^3 -\phi^{-3} = 4,\\\phi^4 -\phi^{-4} = 7,\\\phi^5 -\phi^{-5} = 11\\\cdots$$

It should be evident that as the exponent increases, the second term in each expression will converge to zero, and the entire expression converges to a number.