Fibonacci Number greater than Golden Section to Power less Two
Theorem
For all $n \in \N_{\ge 2}$:
- $F_n \ge \phi^{n - 2}$
where:
- $F_n$ is the $n$th Fibonacci number
- $\phi$ is the golden section: $\phi = \dfrac {1 + \sqrt 5} 2$
Proof
The proof proceeds by induction.
For all $n \in \N_{\ge 2}$, let $\map P n$ be the proposition:
- $F_n \ge \phi^{n - 2}$
Basis for the Induction
$\map P 2$ is true, as this just says:
- $F_2 = 1 = \phi^0 = \phi^{2 - 2}$
It is also necessary to demonstrate $\map P 3$ is true:
- $F_3 = 2 \ge \dfrac {1 + \sqrt 5} 2 = \phi = \phi^{3 - 2}$
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that, if $\map P j$ is true for all $2 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
- $F_j \ge \phi^{j - 2}$ for $2 \le j \le k$
from which it is to be shown that:
- $F_{k + 1} \ge \phi^{k - 1}$
Induction Step
This is the induction step:
As we have already shown $\map P 2$ and $\map P 3$, we just need to prove the result for $k \ge 3$.
\(\ds F_{k + 1}\) | \(=\) | \(\ds F_{k - 1} + F_k\) | Definition of Fibonacci Numbers | |||||||||||
\(\ds \) | \(\ge\) | \(\ds \phi^{k - 3} + \phi^{k - 2}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \phi^{k - 3} \paren {1 + \phi}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \phi^{k - 3} \paren {\phi^2}\) | Square of Golden Mean equals One plus Golden Mean | |||||||||||
\(\ds \) | \(=\) | \(\ds \phi^{k - 1}\) |
So $\map P 2 \land \map P 3 \land \dots \land \map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.
Therefore:
- $\forall n \in \N_{\ge 2}: F_n \ge \phi^{n - 2}$
$\blacksquare$
Also see
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.1$: Mathematical Induction: Exercise $4$