📚 Maclaurin Series Key Points | 麦克劳林展开考点精讲
Maclaurin series is a powerful tool in calculus for representing functions as infinite polynomials, essential for both IB Mathematics: Analysis and Approaches (HL) and Edexcel A Level Further Mathematics. Mastery of its derivation, standard expansions, and applications like approximation and limit evaluation is crucial for top exam performance. This guide covers all key exam points.
麦克劳林级数是微积分中将函数表示为无穷多项式的强大工具，对IB数学分析与方法（HL）和Edexcel A Level进阶数学都至关重要。掌握其推导、常用展开式以及近似计算和极限求值等应用，是取得高分的必备技能。本文精讲所有核心考点。
1. Definition and Formula | 定义与公式
The Maclaurin series of a function f(x) that is infinitely differentiable at x = 0 is the Taylor series centered at 0:
f(x) = f(0) + f ‘(0)x + f ”(0)x²/2! + f ”'(0)x³/3! + … + f⁽ⁿ⁾(0)xⁿ/n! + …
Here f⁽ⁿ⁾(0) denotes the n-th derivative evaluated at x = 0. The series is an infinite sum of terms, each involving a power of x multiplied by the corresponding derivative at zero divided by the factorial of the power index. It is a local representation valid within the radius of convergence.
对于在 x = 0 处无限次可导的函数 f(x)，其麦克劳林级数是以 0 为中心的泰勒级数：
f(x) = f(0) + f ‘(0)x + f ”(0)x²/2! + f ”'(0)x³/3! + … + f⁽ⁿ⁾(0)xⁿ/n! + …
这里 f⁽ⁿ⁾(0) 表示在 x = 0 处的第 n 阶导数值。该级数是无穷项之和，每一项包含 x 的幂次乘以零点处的导数除以该幂次指数的阶乘。它是一个局部表示，在收敛半径内有效。
2. Derivation Process | 推导步骤
To find the Maclaurin series of a given function f(x), follow these steps: first, compute the derivatives f ‘(x), f ”(x), f ”'(x), … up to the required order. Then evaluate each derivative at x = 0 to obtain f(0), f ‘(0), f ”(0), f ”'(0), … . Finally, substitute these values into the series formula f(0) + f ‘(0)x + f ”(0)x²/2! + f ”'(0)x³/3! + … . For instance, if f(x) = eˣ, all derivatives are eˣ, so f⁽ⁿ⁾(0) = 1 for all n, giving the series 1 + x + x²/2! + x³/3! + … .
要求出一个给定函数 f(x) 的麦克劳林级数，需遵循以下步骤：首先求出各阶导数 f ‘(x), f ”(x), f ”'(x), … 直至所需阶数。然后在 x = 0 处计算每一阶导数值，得到 f(0), f ‘(0), f ”(0), f ”'(0), … 。最后将这些值代入级数公式 f(0) + f ‘(0)x + f ”(0)x²/2! + f ”'(0)x³/3! + … 。例如，若 f(x) = eˣ，所有导数均为 eˣ，故对任意 n 有 f⁽ⁿ⁾(0) = 1，得到级数 1 + x + x²/2! + x³/3! + … 。
3. Standard Maclaurin Expansions | 标准麦克劳林展开式
The following table lists the most commonly tested Maclaurin series. Memorising these is essential for efficient problem-solving.
下表列出了最常考的标准麦克劳林级数。牢记这些展开式对高效解题至关重要。
| Function f(x) | Maclaurin Series (first few terms) |
|---|---|
| eˣ | 1 + x + x²/2! + x³/3! + x⁴/4! + … |
| sin x | x – x³/3! + x⁵/5! – x⁷/7! + … |
| cos x | 1 – x²/2! + x⁴/4! – x⁶/6! + … |
| ln(1 + x) | x – x²/2 + x³/3 – x⁴/4 + … (valid for –1 < x ≤ 1) |
| (1 + x)ⁿ | 1 + nx + n(n–1)x²/2! + n(n–1)(n–2)x³/3! + … , |x| < 1 |
Note the alternating signs for sin x and cos x, and the restricted domain for the logarithmic and binomial series. The binomial expansion holds for any real n, not just positive integers, which is a common extension in IB and Edexcel exams.
注意 sin x 和 cos x 的交替符号，以及对数级数和二项式级数的限制定义域。二项式展开对任意实数 n 均成立，而不仅限于正整数，这是IB和Edexcel考试中常见的拓展。
4. Manipulating Known Expansions | 利用已知展开求新展开
Once you have memorised the standard series, you can quickly obtain new ones by substitution, multiplication, differentiation, or integration. For example, to expand e³ˣ, substitute 3x into the series for eˣ: 1 + (3x) + (3x)²/2! + (3x)³/3! + … = 1 + 3x + 9x²/2 + 9x³/2 + … . Similarly, sin(2x) = 2x – (2x)³/3! + … = 2x – 8x³/6 + … = 2x – (4/3)x³ + … . Differentiating the series for sin x term-by-term gives cos x, and integrating cos x gives sin x, which often saves time.
熟记标准级数后，便可通过代换、乘法、微分或积分快速得到新的展开式。例如，要展开 e³ˣ，将 3x 代入 eˣ 的级数中：1 + (3x) + (3x)²/2! + (3x)³/3! + … = 1 + 3x + 9x²/2 + 9x³/2 + … 。类似地，sin(2x) = 2x – (2x)³/3! + … = 2x – 8x³/6 + … = 2x – (4/3)x³ + … 。对 sin x 的级数逐项微分可得 cos x 的级数，对 cos x 积分可得 sin x，这常常能节省时间。
5. Approximation and Error Estimation | 近似计算与误差估计
Truncating a Maclaurin series after a few terms yields a polynomial approximation. The accuracy depends on the order of the truncation and the value of x. The error can be bounded using the Lagrange remainder formula: Rₙ(x) = f⁽ⁿ⁺¹⁾(c) xⁿ⁺¹ / (n+1)! for some c between 0 and x. In exams, you may be asked to approximate a function value and find the maximum possible error, or to determine the number of terms needed for a specified accuracy.
截取麦克劳林级数的前几项可得到一个多项式近似值。其精度取决于截断的阶数和 x 的值。误差可用拉格朗日余项公式界定：Rₙ(x) = f⁽ⁿ⁺¹⁾(c) xⁿ⁺¹ / (n+1)!，其中 c 介于 0 与 x 之间。考试中可能要求近似计算函数值并求最大可能误差，或确定达到指定精度所需的项数。
6. Evaluating Limits Using Series | 利用级数计算极限
Maclaurin expansions transform indeterminate forms (0/0, ∞/∞) into polynomial ratios, making limits straightforward. For example, lim_{x→0} (sin x – x)/x³ can be found by substituting sin x = x – x³/3! + … , giving (x – x³/6 + … – x)/x³ = (–x³/6 + …)/x³ → –1/6 as x→0. This method is particularly useful when L’Hôpital’s rule becomes messy due to multiple derivatives.
麦克劳林展开式将不定式（0/0，∞/∞）转化为多项式之比，使极限计算变得简单直接。例如，求 lim_{x→0} (sin x – x)/x³，代入 sin x = x – x³/3! + … 得 (x – x³/6 + … – x)/x³ = (–x³/6 + …)/x³ → –1/6。当洛必达法则因多次求导而变得繁琐时，这一方法尤为有效。
7. Maclaurin Series for Composite Functions | 复合函数的麦克劳林级数
When a function is a product, such as eˣ sin x, you can multiply the respective series up to the desired power. To find the expansion up to x⁴, multiply (1 + x + x²/2 + x³/6 + x⁴/24) by (x – x³/6) and collect like terms. Another approach is to use known standard series and algebraic manipulation. For composite functions like √(1 + sin x), it is often easier to compute derivatives directly, but substitution inside a series may also work if the inner function is small near 0.
若函数为乘积形式，如 eˣ sin x，可将两个函数的级数相乘至所需幂次。例如，求 x⁴ 以内的展开，将 (1 + x + x²/2 + x³/6 + x⁴/24) 与 (x – x³/6) 相乘并合并同类项。另一种方法是利用已知标准级数和代数运算。对于 √(1 + sin x) 这类复合函数，直接求导往往更简便，但若内层函数在 0 附近很小，也可在级数中代入。
8. Solving Differential Equations | 解微分方程
Maclaurin series can solve differential equations by assuming a power series solution y = Σ aₙ xⁿ, substituting into the equation, and equating coefficients. This is particularly useful for equations that lack closed-form solutions. For example, given y’ = y with y(0) = 1, assume y = a₀ + a₁ x + a₂ x² + … . Then y’ = a₁ + 2a₂ x + 3a₃ x² + … . Substituting gives a₁ + 2a₂ x + … = a₀ + a₁ x + … . With a₀ = 1, we obtain a₁ = 1, a₂ = 1/2, a₃ = 1/6, etc., recovering eˣ.
麦克劳林级数可用于解微分方程：假设一个幂级数解 y = Σ aₙ xⁿ，代入方程并比较系数。这对没有封闭形式解的情形特别有用。例如，对 y’ = y 且 y(0) = 1，假设 y = a₀ + a₁ x + a₂ x² + … ，则 y’ = a₁ + 2a₂ x + 3a₃ x² + … 。代入得 a₁ + 2a₂ x + … = a₀ + a₁ x + … 。由 a₀ = 1 得 a₁ = 1, a₂ = 1/2, a₃ = 1/6 等，恰好恢复 eˣ 的级数。
9. Interval of Convergence | 收敛区间
The Maclaurin series may not converge for all x. The interval of convergence is found using the ratio test. For a series Σ aₙ xⁿ, compute lim_{n→∞} |aₙ₊₁ / aₙ| ⋅ |x| = L. The series converges absolutely when L < 1, giving the radius of convergence R. The endpoints must be tested separately. For eˣ, sin x, and cos x, the radius is infinite. For ln(1 + x), the interval is –1 < x ≤ 1. For the binomial (1 + x)ⁿ, it is |x| < 1.
麦克劳林级数并非对所有 x 都收敛。收敛区间可通过比值检验求得。对于级数 Σ aₙ xⁿ，计算 lim_{n→∞} |aₙ₊₁ / aₙ| ⋅ |x| = L。当 L < 1 时级数绝对收敛，从而得到收敛半径 R。端点需单独检验。eˣ、sin x、cos x 的收敛半径为无穷大；ln(1 + x) 的收敛区间为 –1 < x ≤ 1；二项式 (1 + x)ⁿ 的区间为 |x| < 1。
10. Common Mistakes and Exam Tips | 常见错误与应试技巧
Many students forget to divide by factorials, misplace signs in alternating series, or fail to check the validity of the expansion beyond the first few terms. When deriving a series, always compute derivatives carefully and double-check f(0) and its early derivatives. For composite functions, ensure that the substitution lies within the original series’ radius of convergence. Also, in error estimation, use the next term’s maximum value on the given interval, not just the next term at x.
许多学生会忘记除以阶乘、在交替级数中弄错符号，或未检查前几项之外展开式的有效性。在推导级数时，务必仔细计算导数，并复核 f(0) 及其低阶导数。对于复合函数，应确保代换后的变量落在原级数的收敛半径内。此外，在误差估计中，要利用下一项在给定区间上的最大值，而非仅仅在 x 处的值。
11. Worked Example | 真题精讲
Find the Maclaurin series for f(x) = ln(1 + sin x) up to the term in x⁴.
求 f(x) = ln(1 + sin x) 的麦克劳林级数至 x⁴ 项。
First, use the standard series: sin x = x – x³/6 + … and ln(1 + u) = u – u²/2 + u³/3 – u⁴/4 + … where u = sin x. Substitute u = x – x³/6 into ln(1 + u) and expand, retaining terms up to x⁴. Compute u² = x² – x⁴/3 + … , u³ = x³ + … , u⁴ = x⁴ + … . Then ln(1 + u) = (x – x³/6) – (x² – x⁴/3)/2 + (x³)/3 – x⁴/4 + … = x – x²/2 + x³/6 – x⁴/12 + … . Always combine like terms carefully and discard powers higher than needed.
首先利用标准级数：sin x = x – x³/6 + … 以及 ln(1 + u) = u – u²/2 + u³/3 – u⁴/4 + … ，其中 u = sin x。将 u = x – x³/6 代入 ln(1 + u) 并展开，保留至 x⁴ 项。计算 u² = x² – x⁴/3 + … ，u³ = x³ + … ，u⁴ = x⁴ + … 。则 ln(1 + u) = (x – x³/6) – (x² – x⁴/3)/2 + (x³)/3 – x⁴/4 + … = x – x²/2 + x³/6 – x⁴/12 + … 。务必仔细合并同类项并舍去高于所需幂次的项。
12. Summary and Key Takeaways | 总结与核心要点
Maclaurin series is a fundamental tool that connects derivatives to polynomial approximations. Key revision points: memorise the expansions of eˣ, sin x, cos x, ln(1 + x) and (1 + x)ⁿ; understand term-by-term differentiation and integration; master substitution to generate new series; and practise error estimation using the Lagrange remainder. In IB exams, expect both direct expansion questions and applications such as limit evaluation or differential equations. In Edexcel Further Maths, the emphasis includes series solutions of differential equations and rigorous convergence testing.
麦克劳林级数是连接导数与多项式逼近的基本工具。核心复习要点：熟记 eˣ、sin x、cos x、ln(1 + x) 和 (1 + x)ⁿ 的展开式；理解逐项微分与积分；掌握代换法产生新级数；练习用拉格朗日余项进行误差估计。在IB考试中，会出现直接展开题以及极限求值或微分方程等应用题。在Edexcel进阶数学中，侧重微分方程的级数解和严格的收敛性检验。
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