Fibonacci Number by One Minus Golden Mean plus Fibonacci Number of Index One Less
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Theorem
Let $n \in \Z$.
Then:
- $\hat \phi^n = F_n \hat \phi + F_{n - 1}$
where:
- $F_n$ denotes the $n$th Fibonacci number
- $\hat \phi$ denotes the $1$ minus the golden mean:
- $\hat \phi := 1 - \phi$
Proof
\(\ds F_n \hat \phi + F_{n - 1}\) | \(=\) | \(\ds F_n \paren {-\dfrac 1 \phi} + F_{n - 1}\) | Reciprocal Form of One Minus Golden Mean | |||||||||||
\(\ds \) | \(=\) | \(\ds -\dfrac 1 \phi \paren {F_n - \phi F_{n - 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\dfrac 1 \phi \paren {\paren {-1}^{n + 1} F_{-n} - \phi \paren {-1}^n F_{-\paren {n - 1} } }\) | Fibonacci Number with Negative Index | |||||||||||
\(\ds \) | \(=\) | \(\ds -\dfrac 1 \phi \paren {\paren {-1}^{n + 1} F_{-n} + \phi \paren {-1}^{n + 1} F_{-\paren {n - 1} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{n + 1} \paren {-\dfrac 1 \phi \paren {F_{-n} + \phi F_{-n + 1} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{n + 1} \paren {-\dfrac 1 \phi } \paren {\phi^{-n + 1} }\) | Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^n \paren {\dfrac 1 {\phi^n} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-\dfrac 1 \phi}^n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \hat \phi^n\) |
$\blacksquare$
Also see
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.8$: Fibonacci Numbers: Exercise $11$