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The Fibonacci generating function and Binet formula over $Q (\sqrt{5})$

Example

Let $(F_{n})$ be the Fibonacci sequence and put

$ϕ = \frac{1 + \sqrt{5}}{2}, \hat{ϕ} = \frac{1 - \sqrt{5}}{2} .$

Then, in $Q ⟦ x ⟧$ ,

$\sum_{n \geq 0} F_{n} x^{n} = \frac{x}{1 - x - x^{2}},$

and, in the splitting field $Q (\sqrt{5})$ ,

$F_{n} = \frac{ϕ^{n} - {\hat{ϕ}}^{n}}{\sqrt{5}} (n \geq 0) .$

Facts & Assumptions

Given: The Fibonacci initial values and recurrence.

[L1]

The Fibonacci sequence satisfies $F_{0} = 0$ , $F_{1} = 1$ , and $F_{n + 2} = F_{n + 1} + F_{n}$ (The Fibonacci sequence $F_{0} = 0, F_{1} = 1$ and Lucas sequence $L_{0} = 2, L_{1} = 1$ ).

[L2]

Multiplication by the reciprocal recurrence denominator converts a recurrence into its finite numerator (A coefficient sequence is eventually linearly recurrent if and only if its formal generating function is rational).

[L3]

Over a characteristic-zero splitting field, distinct characteristic roots give a unique linear combination of their powers (Over a named splitting field in characteristic zero, repeated characteristic roots give polynomial-times-exponential closed forms).

[L4]

The factors $t - λ$ of the characteristic polynomial correspond to the factors $1 - λ x$ of the reciprocal denominator (Reciprocal-root convention: $χ (t) = \prod_{i} (t - λ_{i})^{m_{i}}$ corresponds to $Q (x) = \prod_{i} (1 - λ_{i} x)^{m_{i}}$ ).

Verification

technique · direct

1.1

If $F (x) = \sum_{n \geq 0} F_{n} x^{n}$ , coefficient extraction using [L1] gives $(1 - x - x^{2}) F (x) = x$ ; [L2] therefore gives the displayed rational generating function.

givenL1L2algebra

1.2

The polynomial $t^{2} - t - 1$ factors as $(t - ϕ) (t - \hat{ϕ})$ in $Q (\sqrt{5}) [t]$ , in agreement with [L4].

L4algebra

2.1

By [L3], $F_{n} = A ϕ^{n} + B {\hat{ϕ}}^{n}$ . The equations $A + B = F_{0} = 0$ and $A ϕ + B \hat{ϕ} = F_{1} = 1$ give $A = 1 / \sqrt{5}$ and $B = - 1 / \sqrt{5}$ .

step 1.2L1L3algebra

3.1

Substitution in step 2.1 proves Binet's formula, including $n = 0$ and $n = 1$ .

step 2.1∎

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 51 results over 9 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

R. P. Stanley, Enumerative Combinatorics, vol. 1, 2nd ed., Example 4.1.2 (standard reference, not scraped)

The Fibonacci generating function and Binet formula over Q(5)Q(5​)