# Binet Form

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## Contents

## Theorem

Let $m \in \R$.

Define:

\(\displaystyle \Delta\) | \(=\) | \(\displaystyle \sqrt {m^2 + 4}\) | |||||||||||

\(\displaystyle \alpha\) | \(=\) | \(\displaystyle \frac {m + \Delta} 2\) | |||||||||||

\(\displaystyle \beta\) | \(=\) | \(\displaystyle \frac {m - \Delta} 2\) |

### First Form

The recursive sequence:

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

where:

\(\displaystyle U_0\) | \(=\) | \(\displaystyle 0\) | |||||||||||

\(\displaystyle U_1\) | \(=\) | \(\displaystyle 1\) |

has the closed-form solution:

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

### Second Form

The recursive sequence:

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

where:

\(\displaystyle V_0\) | \(=\) | \(\displaystyle 2\) | |||||||||||

\(\displaystyle V_1\) | \(=\) | \(\displaystyle m\) |

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:

\(\displaystyle U_0 + U_2\) | \(=\) | \(\displaystyle 0 + m U_1 + U_0\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle m \times 1 + 0\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle m\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle V_1\) | From definition | ||||||||||

\(\displaystyle U_1 + U_3\) | \(=\) | \(\displaystyle 1 + m U_2 + U_1\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle m \times m + 2\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle m \times V_1 + V_0\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle V_2\) | 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:

\(\displaystyle V_{k + 2}\) | \(=\) | \(\displaystyle m V_{k + 1} + V_k\) | Definition of the recursive sequence | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle m \paren {U_k + U_{k + 2} } + \paren {U_{k - 1} + U_{k + 1} }\) | Induction Hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {m U_k + U_{k - 1} } + \paren {m U_{k + 2} + U_{k + 1} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle U_{k + 1} + U_{k + 3}\) |

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.

## Sources

- Weisstein, Eric W. "Binet Forms." From
*MathWorld*--A Wolfram Web Resource. http://mathworld.wolfram.com/BinetForms.html