Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less/Negative Index
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Theorem
Let $n \in \Z_{\le 0}$.
Then:
- $\phi^n = F_n \phi + F_{n - 1}$
where:
- $F_n$ denotes the $n$th Fibonacci number as extended to negative indices
- $\phi$ denotes the golden mean.
Proof
The proof proceeds by induction.
For all $n \in \Z_{\le 0}$, let $\map P n$ be the proposition:
- $\phi^n = F_n \phi + F_{n - 1}$
This can equivalently be expressed as:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
- $\phi^{-n} = F_{-n} \phi + F_{-n - 1}$
$\map P 0$ is the case:
\(\ds F_0 \times \phi + F_{-1}\) | \(=\) | \(\ds F_0 \times \phi + \paren {-1}^0 F_1\) | 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 $\map P 0$ is seen to hold.
Basis for the Induction
$\map P 1$ is the case:
\(\ds F_{-1} \times \phi + F_{-2}\) | \(=\) | \(\ds \paren {-1}^0 F_1 \times \phi + \paren {-1}^{-1} F_2\) | Fibonacci Number with Negative Index | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 \times \phi - F_2\) | Definition of Fibonacci Number $F_1 = 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \phi - 1\) | Definition of Fibonacci Number: $F_2 = 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 \phi\) | Definition 3 of Golden Mean |
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
- $\phi^{-k} = F_{-k} \phi + F_{-k - 1}$
from which it is to be shown that:
- $\phi^{-\paren {k + 1} } = F_{-\paren {k + 1} } \phi + F_{-\paren {k + 1} - 1}$
Induction Step
This is the induction step:
\(\ds F_{-\paren {k + 1} } \phi + F_{-\paren {k + 1} - 1}\) | \(=\) | \(\ds \paren {-1}^{-\paren {k + 1} + 1} F_{k + 1} \phi + \paren {-1}^{-\paren {k + 1} } F_{\paren {k + 1} + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{k + 2} F_{k + 1} \phi - \paren {-1}^{k + 2} F_{k + 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{k + 2} \paren {F_{k + 1} \phi - \paren {F_{k + 1} + F_k} }\) | Definition of Fibonacci Number | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{k + 2} \paren {F_{k + 1} \paren {\phi - 1} - F_k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{k + 2} \paren {F_{k + 1} \phi^{-1} - F_k}\) | Definition 3 of Golden Mean | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{k + 2} F_{k + 1} \phi^{-1} + \paren {-1}^{k + 1} F_k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 \phi \paren {F_{-k - 1} + F_{-k} \phi}\) | Fibonacci Number with Negative Index | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 \phi \paren {\phi^{-k} }\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \phi^{-\paren {k + 1} }\) |
So $\map P k \implies \map P {k + 1}$ and it follows by the Principle of Mathematical Induction that:
- $\forall n \in \Z_{\le 0}: \phi^n = F_n \phi + F_{n - 1}$
$\blacksquare$