Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $n \in \Z$.

Then:

$\phi^n = F_n \phi + F_{n - 1}$

where:

$F_n$ denotes the $n$th Fibonacci number
$\phi$ denotes the golden mean.


Proof

Positive Index

First the result is proved for positive integers.


The proof proceeds by induction.

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

$\phi^n = F_n \phi + F_{n - 1}$


$P \left({0}\right)$ is the case:

\(\displaystyle F_0 \times \phi + F_{-1}\) \(=\) \(\displaystyle F_0 \times \phi + \left({-1}\right)^0 F_1\) Fibonacci Number with Negative Index
\(\displaystyle \) \(=\) \(\displaystyle F_0 \times \phi + 1\) Definition of Fibonacci Number $F_1 = 1$
\(\displaystyle \) \(=\) \(\displaystyle 0 \times \phi + 1\) Definition of Fibonacci Number $F_0 = 0$
\(\displaystyle \) \(=\) \(\displaystyle 1\)
\(\displaystyle \) \(=\) \(\displaystyle \phi^0\)


Thus $P \left({0}\right)$ is seen to hold.


Basis for the Induction

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

\(\displaystyle F_1 \times \phi + F_0\) \(=\) \(\displaystyle F_1 \times \phi\) Definition of Fibonacci Number $F_0 = 0$
\(\displaystyle \) \(=\) \(\displaystyle 1 \times \phi\) Definition of Fibonacci Number $F_1 = 1$
\(\displaystyle \) \(=\) \(\displaystyle \phi^1\)


Thus $P \left({1}\right)$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true.


So this is the induction hypothesis:

$\phi^k = F_k \phi + F_{k - 1}$


from which it is to be shown that:

$\phi^{k + 1} = F_{k + 1} \phi + F_k$


Induction Step

This is the induction step:


\(\displaystyle F_{k + 1} \phi + F_k\) \(=\) \(\displaystyle \left({F_k + F_{k - 1} }\right) \phi + F_k\) Definition of Fibonacci Number
\(\displaystyle \) \(=\) \(\displaystyle F_k \left({1 + \phi}\right) + F_{k - 1} \phi\)
\(\displaystyle \) \(=\) \(\displaystyle F_k \phi^2 + F_{k - 1} \phi\) Square of Golden Mean equals One plus Golden Mean
\(\displaystyle \) \(=\) \(\displaystyle \phi \left({F_k \phi + F_{k - 1} }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \phi \left({\phi^n}\right)\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle \phi^{n + 1}\)


So $P \left({k}\right) \implies P \left({k + 1}\right)$ and it follows by the Principle of Mathematical Induction that:

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

$\Box$


Negative Index

Then the result is extended to negative integers.


The proof proceeds by induction.

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

$\phi^{-n} = F_{-n} \phi + F_{-n - 1}$


$P \left({0}\right)$ is the case:

\(\displaystyle F_0 \times \phi + F_{-1}\) \(=\) \(\displaystyle F_0 \times \phi + \left({-1}\right)^0 F_1\) Fibonacci Number with Negative Index
\(\displaystyle \) \(=\) \(\displaystyle F_0 \times \phi + 1\) Definition of Fibonacci Number $F_1 = 1$
\(\displaystyle \) \(=\) \(\displaystyle 0 \times \phi + 1\) Definition of Fibonacci Number $F_0 = 0$
\(\displaystyle \) \(=\) \(\displaystyle 1\)
\(\displaystyle \) \(=\) \(\displaystyle \phi^0\)


Thus $P \left({0}\right)$ is seen to hold.


Basis for the Induction

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

\(\displaystyle F_{-1} \times \phi + F_{-2}\) \(=\) \(\displaystyle \left({-1}\right)^0 F_1 \times \phi + \left({-1}\right)^{-1} F_2\) Fibonacci Number with Negative Index
\(\displaystyle \) \(=\) \(\displaystyle 1 \times \phi - F_2\) Definition of Fibonacci Number $F_1 = 1$
\(\displaystyle \) \(=\) \(\displaystyle \phi - 1\) Definition of Fibonacci Number: $F_2 = 1$
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 \phi\) Definition 3 of Golden Mean


Thus $P \left({1}\right)$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true.


So this is the induction hypothesis:

$\phi^{-k} = F_{-k} \phi + F_{-k - 1}$


from which it is to be shown that:

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


Induction Step

This is the induction step:


\(\displaystyle F_{-\left({k + 1}\right)} \phi + F_{-\left({k + 1}\right) - 1}\) \(=\) \(\displaystyle \left({-1}\right)^{-\left({k + 1}\right) + 1} F_{k + 1} \phi + \left({-1}\right)^{-\left({k + 1}\right)} F_{\left({k + 1}\right) + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \left({-1}\right)^{k + 2} F_{k + 1} \phi - \left({-1}\right)^{k + 2} F_{k + 2}\)
\(\displaystyle \) \(=\) \(\displaystyle \left({-1}\right)^{k + 2} \left({F_{k + 1} \phi - \left({F_{k + 1} + F_k}\right)}\right)\) Definition of Fibonacci Number
\(\displaystyle \) \(=\) \(\displaystyle \left({-1}\right)^{k + 2} \left({F_{k + 1} \left({\phi - 1}\right) - F_k}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({-1}\right)^{k + 2} \left({F_{k + 1} \phi^{-1} - F_k}\right)\) Definition 3 of Golden Mean
\(\displaystyle \) \(=\) \(\displaystyle \left({-1}\right)^{k + 2} F_{k + 1} \phi^{-1} + \left({-1}\right)^{k + 1} F_k\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 \phi \left({F_{-k - 1} + F_{-k} \phi}\right)\) Fibonacci Number with Negative Index
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 \phi \left({\phi^{-k} }\right)\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle \phi^{-\left({k + 1}\right)}\)


So $P \left({k}\right) \implies P \left({k + 1}\right)$ and it follows by the Principle of Mathematical Induction that:

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

$\Box$


Hence the result is seen to show for both positive and negative integers.

$\blacksquare$


Also see


Sources