Investigation 1 - The Fibonacci Sequence | 探究1：斐波那契数列

📚 Investigation 1 - The Fibonacci Sequence | 探究1：斐波那契数列

In the early 13th century, while pondering the growth of an idealised rabbit population, the Italian mathematician Leonardo of Pisa – known today as Fibonacci – unwittingly set in motion a mathematical exploration that would echo through centuries. What emerged from his deceptively simple puzzle was a sequence of numbers possessing an almost magical blend of simplicity, beauty, and deep interconnectedness with topics as diverse as plant morphology, art, computer algorithms, and continued fractions. This investigation invites you to explore the Fibonacci sequence, to prove its fundamental properties, and to appreciate why it continues to captivate mathematicians and scientists alike.

13 世纪早期，意大利数学家比萨的莱昂纳多——也就是如今我们熟知的斐波那契——在思考一个理想化的兔群增长问题时，无意间开启了一场回荡数个世纪的数学探索。从他那个看似简单的谜题中诞生了一列数，它兼具简洁、优美，且与植物形态学、艺术、计算机算法、连分数等众多领域有着深刻的联系。本探究邀请你一同探索斐波那契数列，证明它的基本性质，并领悟它为何至今仍令数学家与科学家心驰神往。

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1. The Birth of the Sequence | 数列的诞生

Fibonacci's original problem, presented in his 1202 book Liber Abaci, asked how many pairs of rabbits would be produced in a year if one begins with a single newborn pair, assuming that every month each pair produces a new pair, which becomes productive from the second month onwards. Ignoring deaths and other real-world complications, the population after each month forms the pattern: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ... . The crucial observation is that after the first two months, every term is the sum of the two preceding terms.

斐波那契在 1202 年的《计算之书》中提出的原始问题是：假设从一对新生兔子开始，每个月每对兔子都会产下一对新兔子，而新兔子从第二个月起才具备繁殖能力，那么一年后会有多少对兔子？若忽略死亡和其他现实因素，每个月的兔子对数就形成了以下规律：1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, … 。核心的发现是：从第三项起，每一项都等于前两项之和。

Although the sequence was known in Indian mathematics centuries earlier, Fibonacci's work popularised it in Europe, and the numbers were later named in his honour by Édouard Lucas in the 19th century. The sequence offers an ideal playground for mathematical investigation: its definition is elementary, yet the properties that can be unearthed range from elementary arithmetic to deep algebraic structures.

虽然这个数列早在几个世纪前的印度数学中就已出现，但斐波那契的著作使其在欧洲广为人知，19 世纪数学家爱德华·卢卡后来以斐波那契的名字为之命名。这个数列为数学探究提供了一个理想舞台：其定义极其简单，然而从中挖掘出的性质却能跨越初等算术，直抵深刻的代数结构。

The transition from a whimsical rabbit puzzle to a serious object of study illustrates an important aspect of mathematical thinking: patterns that arise from naive modelling often conceal a rich theoretical backbone. By interrogating the Fibonacci sequence, you will learn to move from spotting a pattern to formulating conjectures and, ultimately, to constructing rigorous proofs.

从一道天真的兔子谜题转变为一个严肃的研究对象，体现了数学思维的一个重要方面：从朴素建模中涌现出的规律，往往蕴藏着丰富的理论内核。通过深入探究斐波那契数列，你将学会从发现规律，到提出猜想，最终构建严谨的证明。

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2. Recursive Definition and First Terms | 递推定义与初始项

Formally, the Fibonacci sequence (F_n) is defined by the initial conditions F₁ = 1, F₂ = 1, and the recurrence relation F_n = F_n-1 + F_n-2 for all integers n ≥ 3. Some modern treatments choose F₀ = 0, F₁ more symmetrical. In this investigation we will adopt the standard IB convention of starting with F₁ = 1, F₂ = 1, but we will occasionally refer to the zero‑indexed version for convenience.

严格地，斐波那契数列 (F_n) 由初始条件 F₁ = 1, F₂ = 1 以及递推关系 F_n = F_n-1 + F_n-2（对所有 n ≥ 3）所定义。有些现代处理方式令 F₀ = 0, F₁ = 1，这会使许多公式显得更加对称。本探究采用 IB 常规约定，以 F₁ = 1, F₂ = 1 为起点，但在方便时也会提及从零开始的下标版本。

Let us record the first ten terms to build familiarity:

让我们列出前十项以建立直观感受：

n	1	2	3	4	5	6	7	8	9	10
Fn	1	1	2	3	5	8	13	21	34	55

A glance at the table reveals that the numbers grow rather rapidly; indeed, the sequence exhibits exponential growth, a fact that can be made precise once we have derived a closed‑form expression. The recursive definition is simple enough to be programmed in a spreadsheet or a few lines of code, yet the sequence's behaviour is far from trivial.

浏览上表即可发现，这些数字增长得相当迅速；事实上，这个数列呈指数级增长，这一点在我们推导出闭式表达式后就可以精确描述。递推定义简单到用电子表格或几行代码就能实现，但数列的行为却远非平凡。

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3. Closed Form: Binet's Formula | 闭式表达式：比内公式

ics. Let φ (phi) and ψ (psi) be the two roots of the quadratic equation x² = x + 1, namely

任何用递推方式定义的数列，一个自然的问题就是：第 n 项是否存在显式公式？对斐波那契数列而言答案是肯定的，这就引出了初等数学中最著名的公式之一。令 φ（phi）和 ψ（psi）为二次方程 x² = x + 1 的两个根，即

φ = (1 + √5)/2, ψ = (1 – √5)/2.

Then the nth Fibonacci number can be written as

那么第 n 个斐波那契数便可写为

F_n = (φⁿ – ψⁿ)/√5,  n ≥ 1.

This identity is traditionally called Binet' he famous golden ratio, while ψ ≈ −0.6180339887... is its negative reciprocal. Notice that |ψ| < 1, so the term ψⁿ becomes very small as n grows; consequently F_n is essentially φⁿ/√5 rounded to the nearest integer.

这个恒等式传统上被称为比内公式，不过早在雅克·比内 1843 年发表它之前，亚伯拉罕·棣莫弗和丹尼尔·伯努利就已经知晓。数 φ ≈ 1.6180339887… 就是著名的黄金比例，而 ψ ≈ −0.6180339887… 则是它的负倒数。注意 |ψ| < 1，因此随着 n 增大，项 ψⁿ 变得非常小；所以 F_n 实质上就是 φⁿ/√5 四舍五入到最接近的整数。

The existence of a closed form is surprising because the sequence is defined solely in terms of integers, yet its formula involves irrational numbers. This illustrates a recurring theme in mathematics: stepping into a larger number system often provides tools that dramatically simplify a problem formulated in a smaller one.

闭式表达式的存在令人惊讶，因为这个数列完全由整数定义，但其公式却涉及无理数。这体现了数学中一个反复出现的主题：踏入更大的数系往往会提供强有力的工具，极大地简化原本在一个较小数系中表述的问题。

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4. Proving Binet's Formula by Induction | 用数学归纳法证明比内公式

The standard technique for proving Binet' and then assume the formula holds for n = k and n = k+1 in order to prove it for n = k+2.

证明比内公式的标准技巧是强数学归纳法。由于递推涉及往前看两步，我们必须验证两个基础情形，然后假设公式对 n = k 和 n = k+1 成立，再证明其对 n = k+2 成立。

Base cases. For n = 1: (φ¹ – ψ¹)/√5 = ( (1+√5)/2 – (1–√5)/2 ) / √5 = (√5)/√5 = 1 = F₁. For n = 2: φ² = φ+1, ψ² = ψ+1, so (φ² – ψ²)/√5 = ( (φ+1) – (ψ+1) )/√5 = (φ – ψ)/√5 = 1 = F₂.

基础情形。 对于 n = 1：(φ¹ – ψ¹)/√5 = ( (1+√5)/2 – (1–√5)/2 ) / √5 = (√5)/√5 = 1 = F₁。对于 n = 2：φ² = φ+1，ψ² = ψ+1，故 (φ² – ψ²)/√5 = ( (φ+1) – (ψ+1) )/√5 = (φ – ψ)/√5 = 1 = F₂。

Inductive step. Assume F_k = (φ^k – ψ^k)/√5 and F_k+1 = (φ^k+1 – ψ^k+1)/√5 for some k ≥ 1. Then using the recurrence,

归纳步骤。 假设对某个 k ≥ 1 有 F_k = (φ^k – ψ^k)/√5 和 F_k+1 = (φ^k+1 – ψ^k+1)/√5。那么利用递推，

F_k+2 = F_k+1 + F_k = (φ^k+1 + φ^k – ψ^k+1 – ψ^k)/√5.

Since φ satisfies φ² = φ + 1, we have φ^k+1 + φ^k = φ^k(φ + 1) = φ^k·φ² = φ^k+2. An identical argument yields ψ^k+1 + ψ^k = ψ^k+2. Substituting these gives

由于 φ 满足 φ² = φ + 1，我们有 φ^k+1 + φ^k = φ^k(φ + 1) = φ^k·φ² = φ^k+2。对 ψ 完全相同的推理可得 ψ^k+1 + ψ^k = ψ^k+2。代回即得

F_k+2 = (φ^k+2 – ψ^k+2)/√5,

which completes the induction. This proof is a beautiful illustration of how the algebraic properties of the golden ratio underpin the arithmetic of the Fibonacci numbers.

这就完成了归纳。这一证明优美地展示了黄金比例的代数性质如何支撑起斐波那契数的算术。

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5. The Golden Ratio Connection | 与黄金比例的关联

76... . The values oscillate but quickly settle near a certain limit. That limit is the golden ratio φ.

斐波那契数列最引人注目的视觉特征之一，就是相邻两项比值的表现。计算前几项比值：1/1 = 1, 2/1 = 2, 3/2 = 1.5, 5/3 ≈ 1.666…, 8/5 = 1.6, 13/8 = 1.625, 21/13 ≈ 1.615…, 34/21 ≈ 1.619…, 55/34 ≈ 1.6176… 。这些值上下摆动，但很快就趋近于某个极限，这个极限就是黄金比例 φ。

To justify this rigorously, assume the limit L = lim_n→∞ F_n+1/F_n exists. From the recurrence, F_n+1 = F_n + F_n-1, divide by F_n: F_n+1/F_n = 1 + F_n-1/F_n. If n is large, F_n-1/F_n 1 + 1/L, leading to L² = L + 1, whose positive solution is φ. The formal proof uses Binet's formula and the fact that |ψ/φ| < 1 to show the limit is exact.

为严格论证这一点，假设极限 L = lim_n→∞ F_n+1/F_n 存在。从递推式 F_n+1 = F_n + F_n-1 两边除以 F_n 得 F_n+1/F_n = 1 + F_n-1/F_n Published by TutorHao | IB Mathematics Revision Series | aleveler.com

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