Fibonacci Number of Even Index by Golden Mean Modulo 1
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Theorem
Let $n \in \Z$ be an integer.
Then:
- $F_{2 n} \phi \bmod 1 = 1 - \phi^{-2 n}$
- $F_n$ denotes the $n$th Fibonacci number
- $\phi$ is the golden mean: $\phi = \dfrac {1 + \sqrt 5} 2$
Proof
From definition of$\bmod 1$, the statement above is equivalent to the statement:
- $F_{2 n} \phi - 1 + \phi^{-2 n}$ is an integer
We have:
\(\ds F_{2 n} \phi - 1 + \phi^{-2 n}\) | \(=\) | \(\ds -1 + F_{2 n + 1} - \hat \phi^{2 n} + \phi^{-2 n}\) | Fibonacci Number n+1 Minus Golden Mean by Fibonacci Number n | |||||||||||
\(\ds \) | \(=\) | \(\ds -1 + F_{2 n + 1} - \paren {-1}^{2 n} \phi^{-2 n} + \phi^{-2 n}\) | Golden Mean by One Minus Golden Mean equals Minus 1 | |||||||||||
\(\ds \) | \(=\) | \(\ds -1 + F_{2 n + 1}\) |
which is an integer.
Hence the result.
$\blacksquare$
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 $31$