Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less/Positive Index
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Theorem
Let $n \in \Z_{\ge 0}$.
Then:
- $\phi^n = F_n \phi + F_{n - 1}$
where:
- $F_n$ denotes the $n$th Fibonacci number
- $\phi$ denotes the golden mean.
Proof
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
- $\phi^n = F_n \phi + F_{n - 1}$
$P \left({0}\right)$ is the case:
\(\ds F_0 \times \phi + F_{-1}\) | \(=\) | \(\ds F_0 \times \phi + \left({-1}\right)^0 F_1\) | Fibonacci Number with Negative Index | |||||||||||
\(\ds \) | \(=\) | \(\ds F_0 \times \phi + 1\) | Definition of Fibonacci Number $F_1 = 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 0 \times \phi + 1\) | Definition of Fibonacci Number $F_0 = 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \phi^0\) |
Thus $P \left({0}\right)$ is seen to hold.
Basis for the Induction
$P \left({1}\right)$ is the case:
\(\ds F_1 \times \phi + F_0\) | \(=\) | \(\ds F_1 \times \phi\) | Definition of Fibonacci Number $F_0 = 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 \times \phi\) | Definition of Fibonacci Number $F_1 = 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \phi^1\) |
Thus $P \left({1}\right)$ is seen to hold.
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true.
So this is the induction hypothesis:
- $\phi^k = F_k \phi + F_{k - 1}$
from which it is to be shown that:
- $\phi^{k + 1} = F_{k + 1} \phi + F_k$
Induction Step
This is the induction step:
\(\ds F_{k + 1} \phi + F_k\) | \(=\) | \(\ds \left({F_k + F_{k - 1} }\right) \phi + F_k\) | Definition of Fibonacci Number | |||||||||||
\(\ds \) | \(=\) | \(\ds F_k \left({1 + \phi}\right) + F_{k - 1} \phi\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds F_k \phi^2 + F_{k - 1} \phi\) | Square of Golden Mean equals One plus Golden Mean | |||||||||||
\(\ds \) | \(=\) | \(\ds \phi \left({F_k \phi + F_{k - 1} }\right)\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \phi \left({\phi^n}\right)\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \phi^{n + 1}\) |
So $P \left({k}\right) \implies P \left({k + 1}\right)$ and it follows by the Principle of Mathematical Induction that:
- $\forall n \in \Z_{\ge 0}: \phi^n = F_n \phi + F_{n - 1}$
$\blacksquare$