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:

$\forall n \in \Z_{\le 0}: \phi^n = F_n \phi + F_{n - 1}$

$\blacksquare$