# Binet Form

## Theorem

Let $m \in \R$.

Define:

 $\ds \Delta$ $=$ $\ds \sqrt {m^2 + 4}$ $\ds \alpha$ $=$ $\ds \frac {m + \Delta} 2$ $\ds \beta$ $=$ $\ds \frac {m - \Delta} 2$

### First Form

$U_n = m U_{n - 1} + U_{n - 2}$

where:

 $\ds U_0$ $=$ $\ds 0$ $\ds U_1$ $=$ $\ds 1$

has the closed-form solution:

$U_n = \dfrac {\alpha^n - \beta^n} \Delta$

### Second Form

$V_n = m V_{n - 1} + V_{n - 2}$

where:

 $\ds V_0$ $=$ $\ds 2$ $\ds V_1$ $=$ $\ds m$

has the closed-form solution:

$V_n = \alpha^n + \beta^n$

where $\Delta, \alpha, \beta$ are as for the first form.

## Relation Between First and Second Form

For any given value of $m$:

$U_{n - 1} + U_{n + 1} = V_n$

## Proof

Proof by induction:

Let $\map P n$ be the proposition:

$U_{n - 1} + U_{n + 1} = V_n$

### Basis for the Induction

We have:

 $\ds U_0 + U_2$ $=$ $\ds 0 + m U_1 + U_0$ $\ds$ $=$ $\ds m \times 1 + 0$ $\ds$ $=$ $\ds m$ $\ds$ $=$ $\ds V_1$ From definition $\ds U_1 + U_3$ $=$ $\ds 1 + m U_2 + U_1$ $\ds$ $=$ $\ds m \times m + 2$ $\ds$ $=$ $\ds m \times V_1 + V_0$ $\ds$ $=$ $\ds V_2$ From definition

Therefore $\map P 1$ and $\map P 2$ are true.

This is the basis for the induction.

### Induction Hypothesis

This is our induction hypothesis:

For some $k \in \N_{> 0}$, both $\map P k$ and $\map P {k + 1}$ are true.

That is:

$U_{k - 1} + U_{k + 1} = V_k$
$U_k + U_{k + 2} = V_{k + 1}$

Now we need to show true for $n = k + 2$:

$\map P {k + 2}$ is true.

That is:

$U_{k + 1} + U_{k + 3} = V_{k + 2}$

### Induction Step

This is our induction step:

 $\ds V_{k + 2}$ $=$ $\ds m V_{k + 1} + V_k$ Definition of the recursive sequence $\ds$ $=$ $\ds m \paren {U_k + U_{k + 2} } + \paren {U_{k - 1} + U_{k + 1} }$ Induction Hypothesis $\ds$ $=$ $\ds \paren {m U_k + U_{k - 1} } + \paren {m U_{k + 2} + U_{k + 1} }$ $\ds$ $=$ $\ds U_{k + 1} + U_{k + 3}$

This show that $\map P {k + 2}$ is true.

By principle of mathematical induction, $\map P n$ is true for all $n \in \N _{> 0}$.

$\blacksquare$

## Source of Name

This entry was named for Jacques Philippe Marie Binet.