Recurrence Relations | 递推关系

📚 Recurrence Relations | 递推关系

A recurrence relation is a powerful mathematical tool that defines each term of a sequence by referring to one or more previous terms. At A-Level, mastering recurrence relations allows you to model discrete change, solve problems involving growth and decay, and connect iterative processes to closed-form expressions. This article covers the essential concepts, from generating terms to solving first-order and second-order linear recurrences using the characteristic equation.

递推关系是一种强大的数学工具，它通过参考前一项或多项来定义序列中的每一项。在 A-Level 阶段，掌握递推关系能够让你对离散变化建模，解决增长和衰减问题，并将迭代过程与通项表达式联系起来。本文将涵盖从生成项到利用特征方程求解一阶和二阶线性递推所需的核心知识点。

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1. What is a Recurrence Relation? | 什么是递推关系？

A recurrence relation is an equation that expresses the nth term of a sequence, usually denoted u_n, in terms of earlier terms such as u_n-1, u_n-2, etc. For example, u_n+1 = u_n + 5 defines a sequence where each term is 5 more than the previous one. The relation must be accompanied by an initial condition (or conditions) to produce a unique sequence.

递推关系是一个用前面的项（如 u_n-1、u_n-2 等）来表示序列第 n 项（通常记作 u_n）的方程。例如，u_n+1 = u_n + 5 定义了一个每一项都比前一项大 5 的序列。该关系必须附带初始条件才能生成唯一的序列。

In Edexcel A-Level Mathematics, recurrence relations appear in the context of sequences and series, and they often require you to find the value of a specific term or to derive a closed-form formula for the general term.

在 Edexcel A-Level 数学中，递推关系出现在数列和级数部分，考题通常要求你找出某一特定项的值，或推导出通项的封闭形式公式。

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2. Generating Terms from a Recurrence | 由递推生成项

To generate terms, you simply substitute the known initial value(s) into the recurrence and work forward step by step. Consider the first-order recurrence u_n+1 = 2u_n - 3 with u₁ = 5. Then u₂ = 2×5 - 3 = 7, u₃ = 2×7 - 3 = 11, u₄ = 2×11 - 3 = 19, and so on.

要生成各项，只需将已知的初始值代入递推式并逐步向前计算。考虑一阶递推 u_n+1 = 2u_n - 3 且 u₁ = 5。则 u₂ = 2×5 - 3 = 7，u₃ = 2×7 - 3 = 11，u₄ = 2×11 - 3 = 19，以此类推。

This iterative process is straightforward but becomes inefficient if you need, say, the 50th term. That is why we often look for a closed form – a direct formula for u_n that does not require computing all preceding terms.

这种迭代方法很简单，但如果你需要第 50 项，就会变得低效。正因如此，我们常常寻求封闭形式——即一个无需计算所有前项就能直接求出 u_n 的公式。

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3. Recurrence vs. Closed Form | 递推形式与通项公式

A closed-form expression gives the nth term as a function of n alone, e.g. u_n = 3×2^n-1 + 1. While a recurrence relation describes how to move from one term to the next, a closed form allows instant evaluation. Converting a recurrence into a closed form is a central skill at A-Level, particularly for linear recurrences.

封闭形式将第 n 项表示为仅依赖于 n 的函数，如 u_n = 3×2^n-1 + 1。递推关系描述如何从一项得到下一项，而封闭形式则允许直接求值。将递推转化为封闭形式是 A-Level 的核心技能之一，尤其对于线性递推关系而言。

For arithmetic and geometric sequences, the link is immediate: the recurrence can be written down directly from the definition, and the closed form is already known. For more complex linear recurrences, we use systematic algebraic methods.

对于等差数列和等比数列，这种联系是直接的：从定义可直接写出递推式，且通项公式已熟知。对于更复杂的线性递推关系，我们使用系统的代数方法。

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4. Arithmetic Sequences via Recurrence | 等差数列的递推表示

An arithmetic sequence with first term a and common difference d satisfies the first-order recurrence u_n+1 = u_n + d, u₁ = a. The closed form is the familiar u_n = a + (n-1)d.

首项为 a、公差为 d 的等差数列满足一阶递推 u_n+1 = u_n + d，u₁ = a。其封闭形式即大家熟知的 u_n = a + (n-1)d。

Recognising this pattern allows you to see that any recurrence of the form u_n+1 = u_n + c generates an arithmetic progression, so you can immediately write down the nth term without further work.

识别这一模式后，你就会发现任何形如 u_n+1 = u_n + c 的递推都生成等差数列，因此你能够立刻写出第 n 项，无需额外运算。

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5. Geometric Sequences via Recurrence | 等比数列的递推表示

A geometric sequence with first term a and common ratio r obeys u_n+1 = r × u_n, u₁ = a. The closed form is u_n = a r^n-1.

首项为 a、公比为 r 的等比数列遵循 u_n+1 = r × u_n，u₁ = a。其封闭形式为 u_n = a r^n-1。

This is the simplest multiplicative recurrence, and it often appears as the homogeneous part of more complex relations. If a recurrence contains a multiplicative constant but no additive term, a geometric sequence is hiding inside.

这是最简单的乘法递推，它经常作为更复杂关系的齐次部分出现。如果一个递推关系中包含乘法常数但没有加法项，其中就隐藏着一个等比数列。

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6. First-Order Linear Recurrences | 一阶线性递推关系

A first-order linear recurrence takes the form u_n+1 = a u_n en b = 0 it reduces to a geometric sequence; when a = 1 it becomes an arithmetic sequence.

一阶线性递推关系具有形式 u_n+1 = a u_n + b，其中 a 和 b 为常数且 a ≠ 0。这种关系结合了缩放因子 a 和常数偏移 b。当 b = 0 时，它就退化为等比数列；当 a = 1 时，则变成等差数列。

At A-Level you may be asked to find u_n in closed form. The general approach is to find a particular (constant) solution and then add the complementary (homogeneous) solution. For a ≠ 1, a constant solution exists: let u_n = c, giving c = a c + b, so c = b/(1-a).

在 A-Level 考试中，你可能会被要求求出 u_n 的封闭形式。一般方法是先求出一个特解（常数解），再加上互补（齐次）解。当 a ≠ 1 时，存在常数解：令 u_n = c，得到 c = a c + b，因此 c = b/(1-a)。

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7. Solving First-Order Recurrences | 一阶递推关系求解

To solve u_n+1 = a u_n + b with a ≠ 1, write the general solution as u_n = A a^n-1 + b/(1-a), where A is determined by the initial condition u₁. Substitute n = 1 to find A = u₁ - b/(1-a). Thus the closed form becomes u_n = (u₁ - b/(1-a)) a^n-1 + b/(1-a).

对于 a ≠ 1 的 u_n+1 = a u_n + b，通解可写为 u_n = A a^n-1 + b/(1-a)，其中 A 由初始条件 u₁ 确定。代入 n = 1 得 A = u₁ - b/(1-a)。因此封闭形式为 u_n = (u₁ - b/(1-a)) a^n-1 + b/(1-a)。

Example: Solve u_n+1 = 3u_n - 4, u₁ = 2. Here a=3, b=-4, so b/(1-a) = -4/(1-3) = 2. The closed form is u_n = (2 - 2)×3^n-1 + 2 = 2. Surprisingly, the sequence is constant.

例题：求解 u_n+1 = 3u_n - 4，u₁ = 2。此处 a=3, b=-4，故 b/(1-a) = -4/(1-3) = 2。封闭形式为 u_n = (2-2)×3^n-1 + 2 = 2。令人惊讶的是，该序列是一个常数序列。

When a = 1 the recurrence is arithmetic: u_n+1 = u_n + b, and the closed form is u_n = u₁ + (n-1)b.

当 a = 1 时，递推为等差数列：u_n+1 = u_n + b，通项公式为 u_n = u₁ + (n-1)b。

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8. Second-Order Linear Homogeneous Recurrences | 二阶线性齐次递推关系

A second-order linear homogeneous recurrence has the form u_n+2 = p u_n+1 + q u_n, where p and q are constants, and two initial conditions u₀ and u₁ (or u₁ and u₂) are given. These appear frequently in modelling, from the Fibonacci sequence to oscillation problems.

二阶线性齐次递推关系的形式为 u_n+2 = p u_n+1 + q u_n，其中 p 和 q 为常数，并给定两个初始条件 u₀ 和 u₁（或 u₁ 和 u₂）。这类关系在建模中频繁出现，从斐波那契数列到振荡问题。

To find a closed form we look for solutions of the type u_n = rⁿ. Substituting gives rⁿ⁺² = p rⁿ⁺¹ + q rⁿ, and dividing by rⁿ (r ≠ 0) yields the characteristic equation r² - p r - q = 0.

为了寻求封闭形式，我们寻找 u_n = rⁿ 类型的解。代入得 rⁿ⁺² = p rⁿ⁺¹ + q rⁿ，两边除以 rⁿ（r ≠ 0）得到特征方程 r² - p r - q = 0。

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9. The Characteristic Equation | 特征方程

The characteristic equation is a quadratic whose roots determine the nature of the sequence. For the recurrence u_n+2 = p u_n+1 + q u_n, we solve r² - p r - q = 0. Let the roots be α and β.

特征方程是一个二次方程，其根决定了序列的性质。对于递推 u_n+2 = p u_n+1 + q u_n，我们求解 r² - p r - q = 0。记两根为 α 和 β。

• If α ≠ β and both are real, the general solution is u_n = A αⁿ + B βⁿ.

若 α ≠ β 且均为实数，通解为 u_n = A αⁿ + B βⁿ。

• If α = β (a repeated real root), the general solution becomes u_n = (A + B n) αⁿ.

若 α = β（重实根），通解变为 u_n = (A + B n) αⁿ。

The constants A and B are found by plugging in the first two initial values. Complex roots are not required for the Edexcel A-Level specification but follow similar principles.

常数 A 和 B 通过代入前两个初始值求得。复数根不在 Edexcel A-Level 考纲范围内，但遵循类似的原理。

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10. Distinct Real Roots | 两不等实根情形

lpha; = 2 and β = -3. Then any sequence satisfying the recurrence can be written as u_n = A×2ⁿ + B×(-3)ⁿ. With initial conditions, say u₀ = 5 and u₁ = 1, we set up simultaneous equations:
When n=0: A + B = 5
When n=1: 2A - 3B = 1
Solving gives A = 16/5, B = 9/5, so u_n = (16/5)×2ⁿ + (9/5)×(-3)ⁿ.

假设特征方程给出两个不同的实根，例如 α = 2，β = -3。那么任何满足递推的序列都可写为 u_n = A×2ⁿ + B×(-3)ⁿ。若给定初始条件，如 u₀ = 5，u₁ = 1，我们建立方程组：
n=0 时：A + B = 5
n=1 时：2A - 3B = 1
解之得 A = 16/5，B = 9/5，因此 u_n = (16/5)×2ⁿ + (9/5)×(-3)ⁿ。

This closed form makes it easy to compute any term, or to analyse long-term behaviour (e.g. as n grows, the term with the larger absolute root dominates).

有了这个封闭形式，我们可以轻松计算任意一项，或分析长期行为（例如随着 n 增大，绝对值较大的根对应的项占主导）。

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11. Repeated Roots | 重根情形

When the characteristic equation has a repeated root α (i.e., discriminant zero), the structure changes because simply writing A αⁿ + B αⁿ would collapse into a single constant times αⁿ and would not provide enough flexibility to satisfy two initial conditions. The correct form is u_n = (A + B n) αⁿ.

当特征方程有重根 α（即判别式为零）时，结构会发生变化，因为简单地写为 A αⁿ + B αⁿ 会合并成单个常数乘以 αⁿ，无法提供足够的灵活性去满足两个初始条件。正确的形式是 u_n = (A + B n) αⁿ。

Example: u_n+2 = 6 u_n+1 - 9 u_n. The characteristic equation is r² - 6r + 9 = 0, giving r = 3 (repeated). The general solution is u_n = (A + B n) 3ⁿ. Using u₀ = 2, u₁ = 5, we get A = 2 from n=0, and 3(A + B) = 5 from n=1, so B = -1/3. Hence u_n = (2 - n/3) 3ⁿ.

例题：u_n+2 = 6 u_n+1 - 9 u_n。特征方程为 r² - 6r + 9 = 0，解得 r = 3（重根）。通解为 u_n = (A + B n) 3ⁿ。利用 u₀ = 2, u₁ = 5，由 n=0 得 A = 2，由 n=1 得 3(A+B) = 5，故 B = -1/3。因此 u_n = (2 - n/3) 3ⁿ。

Note that the n factor introduces a polynomial growth that modifies the pure exponential, which can be useful in distinguishing sequences with identical eigenvalues.

请注意，因子 n 引入了多项式增长，修正了纯指数行为，这在区分具有相同特征值的序列时非常有用。

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