Binet Form/First Form
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Theorem
Let $m \in \R$.
Define:
\(\ds \Delta\) | \(=\) | \(\ds \sqrt {m^2 + 4}\) | ||||||||||||
\(\ds \alpha\) | \(=\) | \(\ds \frac {m + \Delta} 2\) | ||||||||||||
\(\ds \beta\) | \(=\) | \(\ds \frac {m - \Delta} 2\) |
The recursive sequence:
- $U_n = m U_{n - 1} + U_{n - 2}$
where:
\(\ds U_0\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds U_1\) | \(=\) | \(\ds 1\) |
has the closed-form solution:
- $U_n = \dfrac {\alpha^n - \beta^n} \Delta$
Proof
Proof by induction:
Let $\map P n$ be the proposition:
- $U_n = \dfrac {\alpha^n - \beta^n} \Delta$
Basis for the Induction
We have:
\(\ds \frac {\alpha^0 - \beta^0} \Delta\) | \(=\) | \(\ds \frac {1 - 1} \Delta\) | Zeroth Power of Real Number equals One | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds U_0\) | From definition | |||||||||||
\(\ds \frac {\alpha^1 - \beta^1} \Delta\) | \(=\) | \(\ds \frac 1 \Delta \paren {\frac {m + \Delta} 2 - \frac {m - \Delta} 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 \Delta \paren \Delta\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds U_1\) | From definition |
Therefore $\map P 0$ and $\map P 1$ are true.
This is the basis for the induction.
Induction Hypothesis
This is our induction hypothesis:
- For some $k \in \N$, both $\map P k$ and $\map P {k + 1}$ are true.
That is:
- $U_k = \dfrac {\alpha^k - \beta^k} \Delta$
- $U_{k + 1} = \dfrac {\alpha^{k + 1} - \beta^{k + 1}} \Delta$
Now we need to show true for $n = k + 2$:
- $\map P {k + 2}$ is true.
That is:
- $U_{k + 2} = \dfrac {\alpha^{k + 2} - \beta^{k + 2}} \Delta$
Induction Step
This is our induction step:
First we notice that:
\(\ds \alpha^2\) | \(=\) | \(\ds \paren {\frac {m + \Delta} 2}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 4 \paren {m^2 + 2 m \Delta + m^2 + 4}\) | Square of Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 4 \paren {2 m^2 + 2 m \Delta + 4}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {m^2 + m \Delta} 2 + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds m \alpha + 1\) |
Similarly:
\(\ds \beta^2\) | \(=\) | \(\ds \paren {\frac {m - \Delta} 2}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 4 \paren {m^2 - 2 m \Delta + m^2 + 4}\) | Square of Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 4 \paren {2 m^2 - 2 m \Delta + 4}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {m^2 - m \Delta} 2 + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds m \beta + 1\) |
Thus:
\(\ds U_{k + 2}\) | \(=\) | \(\ds m U_{k + 1} + U_k\) | Definition of the recursive sequence | |||||||||||
\(\ds \) | \(=\) | \(\ds m \paren {\frac {\alpha^{k + 1} - \beta^{k + 1} } \Delta} + \frac {\alpha^k - \beta^k} \Delta\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 \Delta \paren {\paren {m \alpha + 1} \alpha^k - \paren {m \beta + 1} \beta^k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 \Delta \paren {\alpha^2 \alpha^k - \beta^2 \beta^k}\) | From above | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\alpha^{k + 2} - \beta^{k + 2} } \Delta\) |
This show that $\map P {k + 2}$ is true.
By principle of mathematical induction, $\map P n$ is true for all $n \in \N$.
$\blacksquare$