closed form expression of fibonacci sequence proof

3 min read 17-03-2025

closed form expression of fibonacci sequence proof

The Fibonacci sequence, a series of numbers where each term is the sum of the two preceding ones (0, 1, 1, 2, 3, 5, 8...), has fascinated mathematicians for centuries. Its elegant simplicity belies a rich mathematical structure. This article delves into the derivation and proof of the closed-form expression, also known as Binet's formula, for calculating any Fibonacci number directly, without iteratively computing the preceding terms.

Understanding the Fibonacci Sequence

The Fibonacci sequence is formally defined by the recurrence relation:

F_n = F_n-1 + F_n-2 for n > 1

with initial conditions F₀ = 0 and F₁ = 1.

This recursive definition is computationally inefficient for larger values of 'n'. A closed-form expression provides a much more efficient way to calculate F_n.

Deriving Binet's Formula

Binet's formula provides a direct calculation for the nth Fibonacci number:

F_n = (φⁿ - ψⁿ) / √5

where:

φ = (1 + √5) / 2 ≈ 1.618 (the golden ratio)
ψ = (1 - √5) / 2 ≈ -0.618

This seemingly magical formula arises from solving the recurrence relation using characteristic equations.

1. Characteristic Equation

We assume a solution of the form F_n = rⁿ, where 'r' is a constant. Substituting this into the recurrence relation:

rⁿ = r^n-1 + r^n-2

Dividing by r^n-2 (assuming r ≠ 0):

r² = r + 1

This is the characteristic equation, a quadratic equation.

2. Solving the Characteristic Equation

Solving the quadratic equation r² - r - 1 = 0 using the quadratic formula yields two roots:

r₁ = φ = (1 + √5) / 2 r₂ = ψ = (1 - √5) / 2

3. General Solution

The general solution to the recurrence relation is a linear combination of these roots:

F_n = Aφⁿ + Bψⁿ

where A and B are constants determined by the initial conditions.

4. Applying Initial Conditions

Using the initial conditions F₀ = 0 and F₁ = 1:

F₀ = Aφ⁰ + Bψ⁰ = A + B = 0 => B = -A F₁ = Aφ¹ + Bψ¹ = Aφ - Aψ = A(φ - ψ) = 1

Solving for A:

A = 1 / (φ - ψ) = 1 / √5

Since B = -A, B = -1 / √5

5. Binet's Formula

Substituting the values of A and B into the general solution:

F_n = (1/√5)φⁿ - (1/√5)ψⁿ = (φⁿ - ψⁿ) / √5

This completes the derivation of Binet's formula.

Proof of Binet's Formula using Mathematical Induction

While the derivation shows how Binet's formula arises, a rigorous proof requires mathematical induction.

Base Cases:

For n = 0: (φ⁰ - ψ⁰) / √5 = (1 - 1) / √5 = 0 = F₀
For n = 1: (φ¹ - ψ¹) / √5 = (φ - ψ) / √5 = √5 / √5 = 1 = F₁

Inductive Hypothesis:

Assume Binet's formula holds true for n = k and n = k-1. That is:

F_k = (φ^k - ψ^k) / √5 F_k-1 = (φ^k-1 - ψ^k-1) / √5

Inductive Step:

We need to show that the formula also holds for n = k+1:

F_k+1 = F_k + F_k-1 (by the definition of the Fibonacci sequence)

Substituting the inductive hypothesis:

F_k+1 = [(φ^k - ψ^k) / √5] + [(φ^k-1 - ψ^k-1) / √5]

F_k+1 = [φ^k + φ^k-1 - (ψ^k + ψ^k-1)] / √5

Since φ² = φ + 1 and ψ² = ψ + 1 (from the characteristic equation), we can rewrite:

φ^k + φ^k-1 = φ^k-1(φ + 1) = φ^k-1φ² = φ^k+1 ψ^k + ψ^k-1 = ψ^k-1(ψ + 1) = ψ^k-1ψ² = ψ^k+1

Therefore:

F_k+1 = (φ^k+1 - ψ^k+1) / √5

This shows that if the formula holds for n = k and n = k-1, it also holds for n = k+1. Combined with the base cases, this completes the proof by mathematical induction.

Conclusion

Binet's formula provides a powerful and elegant closed-form expression for calculating Fibonacci numbers. Its derivation, based on solving the characteristic equation of the recurrence relation, and its proof through mathematical induction, demonstrate the beauty and rigor of mathematical analysis. This formula is essential for various applications, including computer science, finance, and nature modeling, offering a significantly faster method compared to the recursive definition, especially for larger values of 'n'.