Relation between $m$th Fibonacci number and Golden Ratio

Question

Can anyone tell me how to interpret the following expression $F_m\sim\phi^m$?

EDIT:

The following answer was where I have seen this notation.

Answer 1

It essentially implies that the $m^{th}$ Fibonacci number is close-ish to the $m^{th}$ power of the golden ratio $\phi$, and becomes even closer as $m$ grows without bound. A sort of asymptotic equivalence, if you will. I will look at this in two respects - a more intuitive and a more formal approach.

---

For a bit of a hand-wave-y, intuitive approach, if $[x]$ represents the function which rounds $x$ to the nearest integer, we know that

$$F_n = \left[ \frac{1}{\sqrt 5} \phi^n \right]$$

Notably, if anecdotally, for the Lucas numbers, a related sequence, we have

$$L_n = \left[ \phi^n \right]$$

We can (informally!) look at $\sim$ as meaning "close to" or "on the order of" (as in the terms that make the most "contribution" to the value of the functions differ by a constant multiple as the variable becomes larger and larger). Then using the rounding relation above, the result is clear: $F_n \sim \phi^n$ since $F_n$ is just a rounding of $\phi^n/\sqrt 5$.

It is fair to ask where that formula comes from. It is essentially tied to Binet's formula (a derivation), a closed-form relation for the Fibonacci numbers:

$$F_n = \frac{\phi^n - \bar{\phi}^n}{\sqrt 5}$$

where $\bar \phi$ denotes the conjugate of the golden ratio, flipping the sign on the square root, i.e. $\bar \phi = (1 - \sqrt 5)/2$.

Notably, $\bar \phi \approx -0.618$, meaning it has magnitude less than $1$, so as $n$ grows, $\bar{\phi}^n$ shrinks to zero. As noted in the comments by Minus One-Twelfth, the rounding relationship then comes in since $| \bar \phi / \sqrt 5 | \approx 0.276 < 1/2$, establishing the first relation.

---

We can also take a more formal approach.

The asymptotic equivalence symbol used - the $\sim$ - has a formal meaning beyond this "kind of close to" hand-waving I've been using, in that $f(x) \sim g(x)$ if

$$\lim_{x\to\infty} \frac{f(x)}{g(x)} = c$$

for some constant $c$ (note: $c$ must be a real number, it cannot be infinity). Of note, the Wikipedia article cited in Toby Mak's answer chooses $c=1$ (the manipulation to an arbitrary constant can be achieved simply by replacing $f(x)$ with $c \cdot f(x)$, effectively).

This tweaked definition was pointed out to me by Rócherz in the comments for Minus One-Twelfth's answer (taken essentially from Introduction to Algorithms by Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest and Clifford Stein), and can also be taken as a corollary to the big $\Theta$ notation or big $O$ notation. (An article on these sorts of notations can be found on Wikipedia or elsewhere if you want to delve further into that.)

Anyhow, we consider the limit we want to find:

$$\lim_{n\to\infty} \frac{F_n}{\phi^n}$$

We will verify the definition for $c=1/\sqrt 5$. From Binet's formula, then, we have

$$\frac{F_n}{\phi^n} = \frac{1}{\sqrt 5} \cdot \frac{\phi^n - \bar{\phi}^n}{\phi^n} = \frac{1}{\sqrt 5} \cdot \left(1 - \frac{\bar{\phi}^n}{\phi^n} \right)$$

It is a simple exercise in rationalizing the denominator to show that

$$\frac{\bar{\phi}}{\phi} = \frac{1}{2} \left( \sqrt 5 - 3 \right) \approx -0.382$$

and thus, since the magnitude is less than one,

$$\frac{\bar{\phi}^n}{\phi^n} \overset{n\to\infty}{\longrightarrow} 0$$

As a result, then

$$\lim_{n\to\infty} \frac{F_n}{\phi^n} = \frac{1}{\sqrt 5} \implies F_n \sim \phi^n$$

---

So, in short:

The notation $F_n \sim \phi^n$ in this context makes that $F_n/\phi^n$ is a constant as $n$ grows without bound, specifically as the limit is taken as $n \to \infty$.