Series and Differential Equations | 数列与微分方程

📚 Series and Differential Equations | 数列与微分方程

Series and differential equations are two fundamental pillars of calculus, often appearing as separate topics in the IB Mathematics curriculum. However, they are deeply interconnected: differential equations can be solved using power series, and the resulting series solutions provide insights into the behavior of functions that are difficult to obtain in closed form. This article explores the theory and application of series solutions to differential equations, tailored for IB HL students seeking a deeper understanding and preparation for internal assessments or exam extensions.

数列与微分方程是微积分的两大基石，在IB数学课程中通常被作为独立主题讲授。然而，它们之间存在着深刻的联系：微分方程可以用幂级数求解，而级数解能够揭示难以用封闭形式表达的函数行为。本文探讨微分方程的级数解理论及其应用，旨在帮助IB高等级学生加深理解，并为内部评估或考试拓展做好准备。

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1. Power Series: Definition and Convergence | 幂级数：定义与收敛性

A power series about x = 0 (a Maclaurin series) is an infinite sum of the form Σ a_n xⁿ, where a_n are coefficients. The series converges for |x| < R, where R is called the radius of convergence. For IB HL, understanding how to determine R using the ratio test is essential. For example, the series Σ (xⁿ/n!) converges for all x, so R = ∞.

以 x = 0 为中心的幂级数（麦克劳林级数）是形如 Σ a_n xⁿ 的无穷和，其中 a_n 为系数。该级数在 |x| < R 时收敛，R 称为收敛半径。对于IB高等级，运用比值判别法确定 R 是必要技能。例如，级数 Σ (xⁿ/n!) 对所有 x 都收敛，故 R = ∞。

Σ_n=0^∞ a_n xⁿ

The ratio test states that a series Σ u_n converges if lim |u_n+1/u_n| < 1. For a power series, this limit becomes |x| lim |a_n+1/a_n|, giving R = 1/lim |a_n+1/a_n|, provided the limit exists.

比值判别法指出，若 lim |u_n+1/u_n| < 1，则级数 Σ u_n 收敛。对幂级数，该极限化为 |x| lim |a_n+1/a_n|，从而 R = 1/lim |a_n+1/a_n|（假设极限存在）。

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2. Representing Functions as Power Series | 将函数表示为幂级数

Many familiar functions have power series expansions, such as e^x = Σ xⁿ/n!, sin x = Σ (-1)ⁿ x²ⁿ⁺¹/(2n+1)!, and ln(1+x) = Σ (-1)^n-1 xⁿ/n for |x| < 1. These expansions are derived using Taylor's theorem or repeated differentiation. In IB, students often work with Maclaurin series up to the first few terms and use them for approximations.

许多常见函数都有幂级数展开式，例如 e^x = Σ xⁿ/n!，sin x = Σ (-1)ⁿ x²ⁿ⁺¹/(2n+1)!，以及当 |x| < 1 时 ln(1+x) = Σ (-1)^n-1 xⁿ/n。这些展开式可通过泰勒定理或多次求导得到。在IB课程中，学生通常只使用麦克劳林级数的前几项，并用于近似计算。

e^x = Σ_n=0^∞ xⁿ/n! , sin x = Σ_n=0^∞ (-1)ⁿ x²ⁿ⁺¹/(2n+1)!

Knowing these standard expansions is extremely useful because they can serve as building blocks for solving differential equations or for recognizing a series solution as a known function.

熟悉这些标准展开式极为有用，因为它们可作为求解微分方程的基本构件，或帮助识别级数解是否对应于某个已知函数。

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3. Differentiating and Integrating Power Series | 幂级数的逐项微分与积分

perty is crucial for solving differential equations using series. For instance, if y = Σ a_n xⁿ, then y' = Σ n a_n x^n-1. The radius of convergence remains the same, but the behaviour at the endpoints must be checked separately.

在收敛半径内，幂级数可以逐项微分和积分。这一性质对于用级数求解微分方程至关重要。例如，若 y = Σ a_n xⁿ，则 y' = Σ n a_n x^n-1。收敛半径保持不变，但端点的敛散性需要单独检验。

d/dx [ Σ_n=0^∞ a_n xⁿ ] = Σ_n=1^∞ n a_n x^n-1

Integration is similarly straightforward: ∫ Σ a_n xⁿ dx = C + Σ a_n xⁿ⁺¹/(n+1). Later, when substituting a series into a differential equation, you will often shift indices to align powers of x. Mastering index manipulation is key to success with this method.

积分也同样直接：∫ Σ a_n xⁿ dx = C + Σ a_n xⁿ⁺¹/(n+1)。稍后在将级数代入微分方程时，你常常需要平移求和指标以对齐 x 的幂次。掌握指标操作是成功运用该方法的关键。

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4. A Review of First-Order Differential Equations | 一阶微分方程回顾

f(x)g(y), linear equations dy/dx + P(x)y = Q(x) (solved using an integrating factor μ = e^{∫ P dx}), and exact equations. Many problems can be solved analytically with these techniques. However, when an ODE cannot be solved in closed form, or if the solution is expressed with non-elementary functions, series solutions become invaluable.

在使用级数方法之前，请先回顾一阶常微分方程的标准类型：可分离变量方程 dy/dx = f(x)g(y)，线性方程 dy/dx + P(x)y = Q(x)（使用积分因子 μ = e^{∫ P dx} 求解），以及恰当方程。许多问题可用这些技巧解析求解。然而，当ODE无法获得封闭解，或解由非初等函数表达时，级数解便显得极有价值。

dy/dx + P(x)y = Q(x) , μ(x) = e^{∫ P(x)dx}

In the IB HL syllabus, first-order ODEs and their applications are central. Series solutions extend your tool kit, enabling you to approximate solutions when standard methods fail or become too cumbersome.

在IB HL大纲中，一阶常微分方程及其应用是核心内容。级数解扩展了你的工具包，使你能够在标准方法失效或过于繁琐时对解进行近似。

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5. The Method of Series Solutions for Differential Equations | 微分方程的级数求解法

ve a differential equation using a power series, we assume a solution of the form y(x) = Σ_n=0^∞ a_n xⁿ, substitute y and its derivatives into the equation, and then equate coefficients of like powers of x to obtain recurrence relations for the coefficients a_n. The initial or boundary conditions determine the first few coefficients, and all the rest follow from the recurrence. This method often yields the Maclaurin series of the true solution.

利用幂级数求解微分方程时，假设解具有形式 y(x) = Σ_n=0^∞ a_n xⁿ，将 y 及其导函数代入方程，然后令各个同次幂的系数相等，从而得到系数 a_n 的递推关系式。初始条件或边界条件决定了前几项系数，其余系数均由递推关系生成。该方法常常能直接给出真解的麦克劳林级数。

The general steps are: (1) Write y, y', y'', ... as series in summation notation. (2) Substitute into the ODE. (3) Shift summation indices so that all series involve the same power of x, say xⁿ. (4) Combine the series. (5) Equate coefficients to obtain a recurrence relation. (6) Use initial conditions to find a₀, a₁, etc. (7) Solve the recurrence to find the general coefficient, if possible, and identify the series.

一般步骤为：(1) 将 y, y', y'' 等写为求和符号下的级数。(2) 代入ODE。(3) 平移求和指标，使所有级数都包含相同的 x 幂次，例如 xⁿ。(4) 合并级数。(5) 比较系数，得到递推关系。(6) 利用初始条件求出 a₀、a₁ 等。(7) 如有可能，求解递推关系以找出一般项系数，并识别该级数。

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6. Example 1: Solving dy/dx = y | 实例一：求解 dy/dx = y

Consider the simple ODE y' = y with initial condition y(0) = 1. Let y = Σ a_n xⁿ. Then y' = Σ n a_n x^n-1. Substituting into y' = y gives Σ n a_n x^n-1 = Σ a_n xⁿ coefficients, shift the index on the left: let m = n-1, so the left sum becomes Σ (m+1) a_m+1 x^m. Renaming the dummy index back to n, we have Σ (n+1) a_n+1 xⁿ = Σ a_n xⁿ.

考虑简单ODE y' = y，初始条件 y(0)=1。设 y = Σ a_n xⁿ，则 y' = Σ n a_n x^n-1。代入得 Σ n a_n x^n-1 = Σ a_n xⁿ。为比较系数，将左边的指标平移：令 m = n-1，左边变为 Σ (m+1) a_m+1 x^m。将哑变量重新命名为 n，即有 Σ (n+1) a_n+1 xⁿ = Σ a_n xⁿ。

Equating coefficients of xⁿ on both sides yields (n+1)a_n+1 = a_n, or a_n+1 = a_n/(n+1). With y(0) = a₀ = 1, we obtain a_n = 1/n! for all n. Thus y = Σ_n=0^∞ xⁿ/n! = e^x, confirming the well-known solution.

比较两边 xⁿ 的系数，得 (n+1)a_n+1 = a_n，即 a_n+1 = a_n/(n+1)。由 y(0) = a₀ = 1，可得 a_n = 1/n! 对所有 n 成立。因此 y = Σ_n=0^∞ xⁿ/n! = e^x，验证了众所周知的解。

y = Σ_n=0^∞ xⁿ/n! = e^x

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7. Example 2: Solving dy/dx = x y | 实例二：求解 dy/dx = x y

Now solve y' = x y with y(0) = 1. Substitute y = Σ a_n xⁿ and y' = Σ n a_n x^n-1 to get Σ n a_n x^n-1 = Σ a_n xⁿ⁺¹. Shift the left index: Σ (n+1) a_n+1 xⁿ. The right side is Σ a_n xⁿ⁺¹; let m = n+1, then it becomes Σ a_m-1 x^m, with m starting at 1. Thus for n = 0

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