Fibonacci Number with Negative Index
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:
\(\ds F_{-1}\) | \(=\) | \(\ds F_1 - F_0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 - 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \left({-1}\right)^{1 + 1} F_1\) |
So $P(1)$ is seen to hold.
$P \left({2}\right)$ is the case:
\(\ds F_{-2}\) | \(=\) | \(\ds F_0 - F_{-1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0 - 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \left({-1}\right)^{2 + 1} F_2\) |
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:
\(\ds F_{- \left({k + 1}\right)}\) | \(=\) | \(\ds F_{-\left({k - 1}\right)} - F_{-k}\) | Definition of Fibonacci Number for Negative Index | |||||||||||
\(\ds \) | \(=\) | \(\ds \left({-1}\right)^k F_{k - 1} - \left({-1}\right)^{k + 1} F_k\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \left({-1}\right)^k F_{k - 1} + \left({-1}\right)^k F_k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \left({-1}\right)^k \left({F_{k - 1} + F_k}\right)\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \left({-1}\right)^k \left({F_{k + 1} }\right)\) | Definition of Fibonacci Number | |||||||||||
\(\ds \) | \(=\) | \(\ds \left({-1}\right)^{k + 2} \left({F_{k + 1} }\right)\) |
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:
- $\forall n \in \Z_{>0} : F_{-n} = \left({-1}\right)^{n + 1} F_n$
$\blacksquare$
Sources
- 1957: George Bergman: Number System with an Irrational Base (Math. Mag. Vol. 31, no. 2: pp. 98 – 110) www.jstor.org/stable/3029218
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.8$: Fibonacci Numbers: Exercise $8$