Euler-Binet Formula/Negative Index

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Theorem

Let $n \in \Z_{< 0}$ be a negative integer.

Let $F_n$ be the $n$th Fibonacci number (as extended to negative integers).


Then the Euler-Binet Formula:

$F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$

continues to hold.


In the above:

$\phi$ is the golden mean
$\hat \phi = 1 - \phi = -\dfrac 1 \phi$


Proof

Let $n \in \Z_{> 0}$.

Then:

\(\ds \dfrac {\phi^{-n} - \hat \phi^{-n} } {\sqrt 5}\) \(=\) \(\ds \dfrac 1 {\sqrt 5} \paren {\phi^{-n} - \paren {-\dfrac 1 \phi}^{-n} }\) Definition of $\hat \phi$
\(\ds \) \(=\) \(\ds \dfrac 1 {\sqrt 5} \paren {\dfrac 1 {\phi^n} - \paren {-1}^n \phi^n}\) Exponent Combination Laws: Negative Power
\(\ds \) \(=\) \(\ds \dfrac 1 {\sqrt 5} \paren {\paren {-1}^n \paren {-\dfrac 1 \phi}^n - \paren {-1}^n \phi^n}\)
\(\ds \) \(=\) \(\ds \paren {-1}^n \dfrac 1 {\sqrt 5} \paren {\hat \phi^n - \phi^n}\) Definition of $\hat \phi$
\(\ds \) \(=\) \(\ds \paren {-1}^{n + 1} \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}\)
\(\ds \) \(=\) \(\ds \paren {-1}^{n + 1} F_n\) Euler-Binet Formula for positive $n$
\(\ds \) \(=\) \(\ds F_{-n}\) Fibonacci Number with Negative Index

$\blacksquare$


Sources