Maclaurin Series: A-Level WJEC Maths Key Points | 麦克劳林展开：A-Level WJEC 数学考点精讲

📚 Maclaurin Series: A-Level WJEC Maths Key Points | 麦克劳林展开：A-Level WJEC 数学考点精讲

The Maclaurin series is a cornerstone of A-Level WJEC Mathematics, enabling us to express complicated functions as infinite polynomials. Mastering its derivation, standard expansions, convergence tests, and error estimation is essential for both Pure and Applied topics. This article systematically reviews the key concepts and exam techniques you need to excel.

麦克劳林级数是 A-Level WJEC 数学的基石，它使我们能把复杂函数表示为无穷多项式。掌握其推导、标准展开式、收敛性检验以及误差估计对于纯数学和应用数学都至关重要。本文系统梳理了必须掌握的核心概念与应试技巧。

1. Introduction to Taylor and Maclaurin Series | 泰勒与麦克劳林级数简介

A Taylor series expands a function f(x) about a point x = a as an infinite sum of terms calculated from the derivatives at that point: f(x) = f(a) + f'(a)(x-a) + f”(a)(x-a)²/2! + f”'(a)(x-a)³/3! + … When the expansion is centred at a = 0, it simplifies to the Maclaurin series. This special case is particularly useful for approximating functions near zero and is the focus of the WJEC syllabus.

泰勒级数将函数 f(x) 在点 x = a 附近展开为无穷和，其各项由该点处的导数计算得出：f(x) = f(a) + f'(a)(x-a) + f”(a)(x-a)²/2! + f”'(a)(x-a)³/3! + …。当展开中心取 a = 0 时，就简化为麦克劳林级数。这一特殊情况对于近似零点附近的函数尤为有用，也是 WJEC 考纲的重点。

2. Derivation of the Maclaurin Series | 麦克劳林级数的推导

The Maclaurin series formula is obtained by setting a = 0 in the Taylor expansion. For an infinitely differentiable function f, the series is given by:

f(x) = f(0) + f'(0)x + f”(0)x²/2! + f”'(0)x³/3! + … = Σₙ₌₀∞ f⁽ⁿ⁾(0) xⁿ / n!

The coefficients are determined by evaluating the n-th derivative f⁽ⁿ⁾(0) and dividing by n! . To derive the series for a specific function, you compute its successive derivatives at x = 0, identify a pattern, and write the general term.

将泰勒展开中的 a 设为 0 就得到麦克劳林级数公式。对于无穷可微的函数 f，其级数为：

f(x) = f(0) + f'(0)x + f”(0)x²/2! + f”'(0)x³/3! + … = Σₙ₌₀∞ f⁽ⁿ⁾(0) xⁿ / n!

系数由 n 阶导数 f⁽ⁿ⁾(0) 除以 n! 决定。要推导某个函数的级数，需要计算其在 x = 0 处的逐阶导数，找出规律，并写出通项。

3. Maclaurin Series for eˣ | eˣ 的麦克劳林展开

The exponential function eˣ is the simplest Maclaurin expansion because all derivatives of eˣ are eˣ itself, and at x = 0 they equal 1. Thus f⁽ⁿ⁾(0) = 1 for all n, giving:

eˣ = 1 + x + x²/2! + x³/3! + x⁴/4! + … = Σₙ₌₀∞ xⁿ / n!

This series converges for all real x (radius of convergence R = ∞). It is fundamental for approximations such as e ≈ 1 + 1 + 1/2 + 1/6 + … and for composing with other series, like e⁻ˣ².

指数函数 eˣ 的麦克劳林展开最为简单，因为 eˣ 的所有阶导数都是其本身，且在 x = 0 处均等于 1。因此对所有 n 有 f⁽ⁿ⁾(0) = 1，于是：

eˣ = 1 + x + x²/2! + x³/3! + x⁴/4! + … = Σₙ₌₀∞ xⁿ / n!

该级数对所有实数 x 收敛（收敛半径 R = ∞）。它是近似计算 e ≈ 1 + 1 + 1/2 + 1/6 + … 以及与其他级数复合（如 e⁻ˣ²）的基础。

4. Sine and Cosine Series | 正弦与余弦的级数展开

The derivatives of sin x cycle every four steps: sin x → cos x → −sin x → −cos x → sin x. Evaluating at 0 gives the pattern 0, 1, 0, −1, … . Therefore only odd powers appear:

sin x = x − x³/3! + x⁵/5! − x⁷/7! + … = Σₙ₌₀∞ (−1)ⁿ x²ⁿ⁺¹ / (2n+1)!

Similarly, cos x derivatives yield 1, 0, −1, 0, … at 0, producing only even powers:

cos x = 1 − x²/2! + x⁴/4! − x⁶/6! + … = Σₙ₌₀∞ (−1)ⁿ x²ⁿ / (2n)!

Both series converge for all real x. These expansions justify the small-angle approximations sin x ≈ x and cos x ≈ 1 − x²/2 for |x| ≪ 1.

sin x 的导数每四步循环一次：sin x → cos x → −sin x → −cos x → sin x。在 0 处求值得到规律 0, 1, 0, −1, …。因此只有奇次幂出现：

sin x = x − x³/3! + x⁵/5! − x⁷/7! + … = Σₙ₌₀∞ (−1)ⁿ x²ⁿ⁺¹ / (2n+1)!

类似地，cos x 的导数在 0 处为 1, 0, −1, 0, …，只产生偶次幂：

cos x = 1 − x²/2! + x⁴/4! − x⁶/6! + … = Σₙ₌₀∞ (−1)ⁿ x²ⁿ / (2n)!

这两个级数对所有实数 x 都收敛。这些展开式证明了当 |x| ≪ 1 时的小角度近似 sin x ≈ x 和 cos x ≈ 1 − x²/2。

5. Natural Logarithm ln(1+x) | 自然对数 ln(1+x) 的展开

The function f(x) = ln(1+x) has derivatives f'(x) = (1+x)⁻¹, f”(x) = −(1+x)⁻², f”'(x) = 2(1+x)⁻³, … giving f⁽ⁿ⁾(0) = (−1)ⁿ⁻¹ (n−1)! for n ≥ 1, and f(0)=0. This leads to the alternating series:

ln(1+x) = x − x²/2 + x³/3 − x⁴/4 + … = Σₙ₌₁∞ (−1)ⁿ⁻¹ xⁿ / n

This expansion is valid for −1 < x ≤ 1. Note that at x = 1 it gives the alternating harmonic series for ln 2, a classic test for conditional convergence.

函数 f(x) = ln(1+x) 的导数为 f'(x) = (1+x)⁻¹，f”(x) = −(1+x)⁻²，f”'(x) = 2(1+x)⁻³，… 从而对 n ≥ 1 有 f⁽ⁿ⁾(0) = (−1)ⁿ⁻¹ (n−1)!，且 f(0)=0。由此得到交错级数：

ln(1+x) = x − x²/2 + x³/3 − x⁴/4 + … = Σₙ₌₁∞ (−1)ⁿ⁻¹ xⁿ / n

该展开式的有效区间是 −1 < x ≤ 1。注意当 x = 1 时它给出 ln 2 的交错调和级数，这是条件收敛的经典实例。

6. Binomial Series (1+x)ⁿ | 二项式级数 (1+x)ⁿ

For any real exponent n, the Maclaurin series of (1+x)ⁿ is the binomial expansion:

(1+x)ⁿ = 1 + nx + n(n−1)x²/2! + n(n−1)(n−2)x³/3! + …

The general coefficient is the binomial coefficient ⁿCₖ = n(n−1)…(n−k+1)/k! . When n is a positive integer, the series terminates after (n+1) terms and is valid for all x. For non-integer n, the series is infinite and converges only for |x| < 1 (or x = 1 under certain conditions). Important special cases include √(1+x) = (1+x)^½ and 1/√(1−2xt+t²) for Legendre polynomials.

对于任意实数指数 n，(1+x)ⁿ 的麦克劳林级数就是二项式展开：

(1+x)ⁿ = 1 + nx + n(n−1)x²/2! + n(n−1)(n−2)x³/3! + …

通项系数为二项式系数 ⁿCₖ = n(n−1)…(n−k+1)/k!。当 n 为正整数时，级数在 (n+1) 项后终止，并对所有 x 成立。对于非整数 n，级数为无穷，仅当 |x| < 1 时收敛（或在特定条件下 x = 1）。重要的特例包括 √(1+x) = (1+x)^½ 以及用于勒让德多项式的 1/√(1−2xt+t²)。

7. Interval of Convergence | 收敛区间

Every Maclaurin series has a radius of convergence R such that the series converges for |x| < R and diverges for |x| > R. To find R, the ratio test is commonly used: compute limₙ→∞ |aₙ₊₁ / aₙ|. If the limit is L, then R = 1/L. For the endpoints x = ±R, convergence must be tested separately, often using the alternating series test or p-test. In WJEC exams, you may be asked to determine the interval of convergence for a given series or to state the valid range for a standard expansion.

8. Lagrange Remainder and Error Bound | 拉格朗日余项与误差界

When truncating a Maclaurin series after the term in xⁿ, the error Rₙ(x) is given by Lagrange’s form:

Rₙ(x) = f⁽ⁿ⁺¹⁾(c) xⁿ⁺¹ / (n+1)!

for some c between 0 and x. This provides a bound on the error when approximating f(x) by its n-th degree Maclaurin polynomial. For bounded derivatives, you can find an upper bound for |Rₙ(x)| and hence determine the number of terms needed for a desired accuracy. Typical exam questions ask, for instance, to show that the error in approximating e^x by 1 + x + x²/2 for |x| < 0.1 is less than 0.0002.

当截断麦克劳林级数到 xⁿ 项时，误差 Rₙ(x) 由拉格朗日余项给出：

Rₙ(x) = f⁽ⁿ⁺¹⁾(c) xⁿ⁺¹ / (n+1)!

其中 c 介于 0 与 x 之间。这为用 n 次麦克劳林多项式近似 f(x) 时的误差提供了一个界。对于有界的导数，可求出 |Rₙ(x)| 的上界，从而确定达到所需精度所需的项数。典型的考试题例如：证明当 |x| < 0.1 时，用 1 + x + x²/2 近似 e^x 的误差小于 0.0002。

9. Applications: Approximations and Limits | 应用：近似计算与极限

Maclaurin series are powerful tools for approximating function values and evaluating limits. By substituting a small value into a truncated series, you quickly obtain an approximation. For example, e⁰·¹ ≈ 1 + 0.1 + 0.005 + 0.000166… = 1.10517. For limits, replacing functions with their series often resolves indeterminate forms. Consider limₓ→₀ (eˣ − 1 − x)/x². Substituting eˣ = 1 + x + x²/2 + x³/6 + … yields the limit 1/2. This technique is especially useful when L’Hôpital’s rule is cumbersome.

麦克劳林级数是函数值近似和极限计算的强大工具。将小值代入截断的级数，即可快速获得近似值。例如 e⁰·¹ ≈ 1 + 0.1 + 0.005 + 0.000166… = 1.10517。求解极限时，用级数替换函数常常能消去不定型。考虑 limₓ→₀ (eˣ − 1 − x)/x²，代入 eˣ = 1 + x + x²/2 + x³/6 + … 可得极限为 1/2。当洛必达法则繁琐时，这一技巧尤为实用。

10. Exam Tips and Common Mistakes | 考试技巧与常见错误

Memorise the five standard Maclaurin series: eˣ, sin x, cos x, ln(1+x), (1+x)ⁿ. Know their general terms and ranges of validity. Common pitfalls include: forgetting to check the convergence interval, especially for ln(1+x) and binomial series; misapplying the factorial in the denominator; confusing the sine and cosine patterns; and neglecting to state the Lagrange remainder correctly. In exam solutions, always show the steps of derivative evaluation and pattern identification, and when using the ratio test, clearly state the condition for convergence. Practice composing series, such as substituting 2x into sin x to get sin 2x, or multiplying series to obtain eˣ sin x up to x⁴.

熟记五个标准麦克劳林级数：eˣ, sin x, cos x, ln(1+x), (1+x)ⁿ。掌握它们的通项和有效范围。常见易错点包括：忘记检查收敛区间，特别是 ln(1+x) 和二项式级数；分母阶乘使用不当；混淆正弦和余弦的规律；以及未能正确陈述拉格朗日余项。在考试作答时，务必展示导数求值和规律识别的步骤；使用比值判别法时，要清晰陈述收敛条件。练习级数的复合，例如将 2x 代入 sin x 得到 sin 2x，或相乘级数求得 eˣ sin x 至 x⁴ 项。

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Maclaurin Series: A-Level WJEC Maths Key Points | 麦克劳林展开：A-Level WJEC 数学考点精讲