Fibonacci Number with Negative Index

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Theorem

Let $F_n$ be the $n$th Fibonacci number.


Then:

$\forall n \in \Z_{> 0} : F_{-n} = \left({-1}\right)^{n + 1} F_n$


Proof

From the initial definition of Fibonacci numbers, we have:

$F_0 = 0, F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3$

By definition of the extension of the Fibonacci numbers to negative integers:

$F_n = F_{n + 2} - F_{n - 1}$


The proof proceeds by induction.


For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:

$F_{-n} = \left({-1}\right)^n F_n$


Basis for the Induction

$P \left({1}\right)$ is the case:

\(\displaystyle F_{-1}\) \(=\) \(\displaystyle F_1 - F_0\)
\(\displaystyle \) \(=\) \(\displaystyle 1 - 0\)
\(\displaystyle \) \(=\) \(\displaystyle 1\)
\(\displaystyle \) \(=\) \(\displaystyle \left({-1}\right)^{1 + 1} F_1\)

So $P(1)$ is seen to hold.


$P \left({2}\right)$ is the case:

\(\displaystyle F_{-2}\) \(=\) \(\displaystyle F_0 - F_{-1}\)
\(\displaystyle \) \(=\) \(\displaystyle 0 - 1\)
\(\displaystyle \) \(=\) \(\displaystyle -1\)
\(\displaystyle \) \(=\) \(\displaystyle \left({-1}\right)^{2 + 1} F_2\)

So $P(2)$ is seen to hold.


This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ and $P \left({k-1}\right)$ are true, where $k > 1$, then it logically follows that $P \left({k+1}\right)$ is true.


So this is our induction hypothesis:

$F_{-\left({k - 1}\right)} = \left({-1}\right)^k F_{k - 1}$
$F_{-k} = \left({-1}\right)^{k + 1} F_k$


Then we need to show:

$F_{-\left({k + 1}\right)} = \left({-1}\right)^{k + 2} F_{k + 1}$


Induction Step

This is our induction step:

\(\displaystyle F_{- \left({k + 1}\right)}\) \(=\) \(\displaystyle F_{-\left({k - 1}\right)} - F_{-k}\) Definition of Fibonacci Number for Negative Index
\(\displaystyle \) \(=\) \(\displaystyle \left({-1}\right)^k F_{k - 1} - \left({-1}\right)^{k + 1} F_k\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle \left({-1}\right)^k F_{k - 1} + \left({-1}\right)^k F_k\)
\(\displaystyle \) \(=\) \(\displaystyle \left({-1}\right)^k \left({F_{k - 1} + F_k}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({-1}\right)^k \left({F_{k + 1} }\right)\) Definition of Fibonacci Number
\(\displaystyle \) \(=\) \(\displaystyle \left({-1}\right)^{k + 2} \left({F_{k + 1} }\right)\)

So $P \left({k}\right) \land P \left({k-1}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \Z_{>0} : F_{-n} = \left({-1}\right)^{n + 1} F_n$

$\blacksquare$


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