Honsberger's Identity/Negative Indices

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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 Honsberger's Identity:

$F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$

continues to hold, whether $m$ or $n$ are positive or negative.


Proof

The proof proceeds by induction.

For all $n \in \Z_{\le 0}$, let $\map P n$ be the proposition:

$F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$


This can equivalently be written:

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

$F_{m - n} = F_{m - 1} F_{-n} + F_m F_{-n + 1}$


$\map P 0$ is the case:

\(\ds F_{m - 1} F_0 + F_m F_1\) \(=\) \(\ds F_m\) Definition of Fibonacci Number: $F_0 = 0, F_1 = 1$
\(\ds \) \(=\) \(\ds F_{m + 0}\)

Thus $\map P 0)$ is seen to hold.


Basis for the Induction

$\map P 1$ is the case:

\(\ds F_{m - 1} F_{-1} + F_m F_0\) \(=\) \(\ds F_{m - 1} F_{-1}\) Definition of Fibonacci Number: $F_0 = 0$
\(\ds \) \(=\) \(\ds F_{m - 1} \paren {-1}^0 F_1\) Fibonacci Number with Negative Index
\(\ds \) \(=\) \(\ds F_{m - 1} F_1\)
\(\ds \) \(=\) \(\ds F_{m - 1}\) Definition of Fibonacci Number: $F_1 = 1$


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 $\map P k$ and $\map P {k - 1}$ are true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$F_{m - k} = F_{m - 1} F_{-k} + F_m F_{-k + 1}$

and:

$F_{m - \paren {k - 1} } = F_{m - 1} F_{-\paren {k - 1} } + F_m F_{-\paren {k - 1} + 1}$


from which it is to be shown that:

\(\ds F_{m - \paren {k + 1} }\) \(=\) \(\ds F_{m - 1} F_{-\paren {k + 1} } + F_m F_{-\paren {k + 1} + 1}\)
\(\ds \leadsto \ \ \) \(\ds F_{m - k - 1}\) \(=\) \(\ds F_{m - 1} F_{-k - 1} + F_m F_{-k}\)


Induction Step

First we note that:

\(\ds F_{m - k}\) \(=\) \(\ds F_{m - 1} F_{-k} + F_m F_{-k + 1}\)
\(\ds \leadstoandfrom \ \ \) \(\ds F_{m - k}\) \(=\) \(\ds F_{m - 1} F_{-k} + F_m F_{-\paren {k - 1} }\)

and:

\(\ds F_{m - \paren {k - 1} }\) \(=\) \(\ds F_{m - 1} F_{-\paren {k - 1} } + F_m F_{-\paren {k - 1} + 1}\)
\(\ds \leadstoandfrom \ \ \) \(\ds F_{m - k + 1}\) \(=\) \(\ds F_{m - 1} F_{-k + 1} + F_m F_{-k + 2}\)


This is the induction step:

\(\ds F_{m - k - 1}\) \(=\) \(\ds F_{m - k + 1} - F_{m - k}\) Definition of Fibonacci Number for Negative Index
\(\ds \) \(=\) \(\ds F_{m - \paren {k - 1} } - F_{m - k}\)
\(\ds \) \(=\) \(\ds \paren {F_{m - 1} F_{-\paren {k - 1} } + F_m F_{-\paren {k - 1} + 1} } - \paren {F_{m - 1} F_{-k} + F_m F_{-k + 1} }\) Induction Hypothesis
\(\ds \) \(=\) \(\ds F_{m - 1} \paren {F_{-\paren {k - 1} } - F_{-k} } + F_m \paren {F_{-\paren {k - 1} + 1} - F_{-k + 1} }\)
\(\ds \) \(=\) \(\ds F_{m - 1} \paren {F_{-k + 1} - F_{-k} } + F_m \paren {F_{-k + 2} - F_{-k + 1} }\)
\(\ds \) \(=\) \(\ds F_{m - 1} F_{-k - 1} + F_m F_{-k}\) Definition of Fibonacci Number for Negative Index

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


Therefore:

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


Sources

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