Answer
1. **\(\phi\) satisfies** \(\phi^2 - \phi - 1=0\).
2. **The two solutions** are \(\phi_1 = \frac{1 + \sqrt{5}}{2}\) and \(\phi_2 = \frac{1 - \sqrt{5}}{2}\).
3. **Fibonacci closed‐form**:
\[
F_n = \frac{\phi_1^n - \phi_2^n}{\,\phi_1 - \phi_2\,},
\quad\text{where}\;\phi_1 - \phi_2 = \sqrt{5}.
\]
Work Step by Step
Below is a concise discussion of the golden‐ratio quadratic equation and its connection to the **closed‐form (Binet) formula** for the Fibonacci sequence.
---
## 1. The Golden‐Ratio Quadratic Equation
We define
\[
\phi \;=\;\frac{1 + \sqrt{5}}{2}.
\]
### (a) Show that \(\phi\) satisfies \(\phi^2 - \phi - 1 = 0\)
Simply compute \(\phi^2\):
\[
\phi^2
= \Bigl(\frac{1 + \sqrt{5}}{2}\Bigr)^2
= \frac{(1 + \sqrt{5})^2}{4}
= \frac{1 + 2\sqrt{5} + 5}{4}
= \frac{6 + 2\sqrt{5}}{4}
= \frac{2(3 + \sqrt{5})}{4}
= \frac{3 + \sqrt{5}}{2}.
\]
Hence
\[
\phi^2 - \phi
= \frac{3 + \sqrt{5}}{2} \;-\; \frac{1 + \sqrt{5}}{2}
= \frac{(3 + \sqrt{5}) - (1 + \sqrt{5})}{2}
= \frac{2}{2}
= 1.
\]
Thus \(\phi^2 - \phi - 1 = 0.\)
---
## 2. The Two Solutions of \(t^2 - t - 1 = 0\)
Solving \(t^2 - t - 1=0\) by the quadratic formula:
\[
t = \frac{1 \pm \sqrt{1 + 4}}{2}
= \frac{1 \pm \sqrt{5}}{2}.
\]
Hence the two roots are
\[
\phi_1 = \frac{1 + \sqrt{5}}{2},
\quad
\phi_2 = \frac{1 - \sqrt{5}}{2}.
\]
These are often called
- \(\phi = \phi_1 = \tfrac{1 + \sqrt{5}}{2}\) (the “golden ratio”), and
- \(\hat{\phi} = \phi_2 = \tfrac{1 - \sqrt{5}}{2}\) (sometimes the “conjugate” golden ratio).
---
## 3. The Fibonacci Closed‐Form (Binet’s Formula)
Recall the Fibonacci sequence \(\{F_n\}\) defined by
\[
F_0 = 0,\quad F_1 = 1,\quad
F_{n} = F_{n-1} + F_{n-2}\quad(n\ge2).
\]
One can show (e.g., by solving the characteristic equation \(t^2 - t - 1=0\)) that
\[
F_n
= \frac{\phi^n - \bigl(\tfrac{1 - \sqrt{5}}{2}\bigr)^n}{\sqrt{5}},
\]
or more symmetrically,
\[
\boxed{
F_n
= \frac{\phi_1^n - \phi_2^n}{\,\phi_1 - \phi_2\,},
}
\]
where \(\phi_1 - \phi_2 = \sqrt{5}\). Substituting \(\phi_1 = \tfrac{1 + \sqrt{5}}{2}\) and \(\phi_2 = \tfrac{1 - \sqrt{5}}{2}\) yields the usual “Binet” form:
\[
F_n
= \frac{\bigl(\tfrac{1 + \sqrt{5}}{2}\bigr)^n
- \bigl(\tfrac{1 - \sqrt{5}}{2}\bigr)^n}{\sqrt{5}}.
\]
That is the closed‐form expression for the \(n\)th Fibonacci number in terms of \(\phi_1\) and \(\phi_2\).
