13

\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

Darío A. Gutiérrez
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Alaa Nasr
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  • Similar to https://math.stackexchange.com/questions/2769805/how-does-fibonacci-recurrence-simplify/2810791 – lhf Jul 12 '18 at 01:56
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    Nice observation! Well done! – lhf Jul 12 '18 at 01:58
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    If you really want your mind blown, try dividing each number by sqrt(5)... – CJ Dennis Jul 12 '18 at 04:41
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    You might also count the occurrences of number-of-leading $0$ resp $9$ ($2 \times $ 0or 9), ($5 \times $ 00or 99), ... and so on and look at the sequence of that counts. Also an interesting statistic - perhaps motivating enough to look further ... – Gottfried Helms Jul 14 '18 at 08:13

3 Answers3

26

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.

orlp
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  • See also https://en.wikipedia.org/wiki/Lucas_number#Relationship_to_Fibonacci_numbers – lhf Jul 12 '18 at 01:58
11

$\phi$ 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.

lulu
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  • "From the quadratic, we see that... satisfies the Fibonacci recursion" - how does this follow so easily/simply? – Itai Jul 16 '18 at 05:53
  • @Itai Ah, because $x^2-x-1$ is the characteristic polynomial of the recursion $a_n=a_{n-1}+a_{n-2}$. – lulu Jul 16 '18 at 10:17
  • So in general, if $z_1,\ldots,z_k$ are the roots of the char. polynomial of a recursion relation then the sequence $z_1^{s_1n} + \ldots + z_k^{s_kn}$ satisfies the recursion (where $s_i=\pm 1$ s.t. conjugate roots have opposite signs)? Or....? – Itai Jul 17 '18 at 07:31
  • @Itai Yes. That is always true. If you want to be sure that you have the complete solution space to the relevant recursion then you need to handle the case of multiple roots (if, say, $\lambda$ is a double root then $n\lambda^n$ will also satisfy the recursion, and so on). – lulu Jul 17 '18 at 09:55
1

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.

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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RayDansh
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