Maclaurin Expansion | 麦克劳林展开 考点精讲

📚 Maclaurin Expansion | 麦克劳林展开 考点精讲

The Maclaurin series is a cornerstone of calculus, transforming complicated functions into infinite polynomials centred at zero. In the OCR IGCSE Additional Mathematics syllabus, mastering Maclaurin expansions enables you to approximate functions, evaluate limits, and solve differential equations with elegance. This article unpacks every key concept, formula, and trick you need to excel in exam questions.

麦克劳林级数是微积分的基石，它将复杂函数转化为以零为中心的无穷多项式。在 OCR IGCSE 附加数学大纲中，掌握麦克劳林展开不仅能帮你优雅地逼近函数、求极限，还能解微分方程。本文梳理了每个核心概念、公式与解题技巧，助你在考试中脱颖而出。

1. What Is a Maclaurin Series? | 什么是麦克劳林级数？

A Maclaurin series is a Taylor series expansion of a function f(x) about x = 0. It expresses the function as an infinite sum of terms built from the function's derivatives at zero, each multiplied by a power of x and divided by the corresponding factorial. If the series converges, it exactly equals f(x) for values within the interval of convergence.

麦克劳林级数是函数 f(x) 在 x = 0 处的泰勒级数展开。它将函数表示成无穷多项之和，每一项由函数在零点的各阶导数、x 的幂次以及对应的阶乘构造而成。若级数收敛，则在收敛区间内它精确等于原函数。

Key idea: we use polynomial approximations to mimic the behaviour of a smooth function near the origin. The more terms we take, the more accurate the approximation becomes.

核心思想：用多项式在原点附近模仿光滑函数的性态。取的项数越多，近似越精确。

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2. The General Maclaurin Formula | 麦克劳林一般公式

The standard Maclaurin series for a function f(x) is:

麦克劳林级数的标准形式为：

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, and n! = 1 × 2 × … × n. The series can be written compactly using sigma notation: ∑[f⁽ⁿ⁾(0) xⁿ / n!] from n=0 to ∞, where the 0-th derivative is the function itself.

其中 f⁽ⁿ⁾(0) 表示在 x = 0 处的 n 阶导数，n! = 1 × 2 × … × n。级数可紧凑地记为求和符号 ∑[f⁽ⁿ⁾(0) xⁿ / n!]（n 从 0 到 ∞），这里 0 阶导就是函数自身。

To build an expansion, you must repeatedly differentiate the function, substitute x 0, and divide by the appropriate factorial. This systematic process is the exam favourite.

要构造展开式，必须反复求导、代入 x = 0，再除以相应的阶乘。这一系统流程是考试的常客。

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

The exponential function eˣ is the simplest Maclaurin expansion because all its derivatives are eˣ, and e⁰ = 1. Therefore, f⁽ⁿ⁾(0) = 1 for every n. Substituting into the general formula gives:

指数函数 eˣ 的麦克劳林展开最为简单，因为它的所有阶导仍是 eˣ，且 e⁰ = 1。故对每个 n 都有 f⁽ⁿ⁾(0) = 1。代入一般公式得：

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

This series converges for all real x. You can use it to approximate e, e², or any exponential by truncating after a few terms.

此级数对所有实数 x 收敛。你可以用它截断前几项来近似计算 e、e² 或任意指数值。

+ 0.5²/2 + 0.5³/6 = 1 + 0.5 + 0.125 + 0.020833 ≈ 1.64583, close to the true value 1.64872.

示例：估算 e⁰·⁵，取 x = 0.5：1 + 0.5 + 0.25/2 + 0.125/6 ≈ 1.64583，接近于真值 1.64872。

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4. Maclaurin Series for sin x and cos x | sin x 与 cos x 的麦克劳林展开

For sin x, the derivatives cycle: f(x)=sin x, f'(x)=cos x, f''(x)=−sin x, f'''(x)=−cos x, and back to sin x. Evaluating at 0 gives: f(0)=0, f'(0)=1, f''(0)=0, f'''=−1, f⁽⁴⁾(0)=0, etc. The resulting series contains only odd powers with alternating signs:

对 sin x 而言，导数循环：sin x → cos x → −sin x → −cos x → sin x。在 0 点取值得到：f(0)=0, f'(0)=1, f''(0)=0, f'''(0)=−1, f⁽⁴⁾(0)=0，以此类推。得到的级数仅含奇次幂，且正负交替：

sin x = x − x³/3! + x⁵/5! − x⁷/7! + …

For cos x, the derivative cycle starts: f(0)=1, f'(0)=0, f''(0)=−1, f'''(0)=0, f⁽⁴⁾(0)=1, etc. Hence, only even powers appear:

cos x 的导数循环：f(0)=1, f'(0)=0, f''(0)=−1, f'''(0)=0, f⁽⁴⁾(0)=1，等等。所以只出现偶次幂：

cos x = 1 − x²/2! + x⁴/4! − x⁶/6! + …

Both series converge for all real x. These expansions are extremely useful for small-angle approximations in physics and engineering problems on the IGCSE paper.

这两个级数对所有实数 x 收敛。在 IGCSE 试卷的物理和工程情境中，它们常用于小角度近似。

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5. Maclaurin Series for ln(1 + x) | ln(1+x) 的麦克劳林展开

For the natural logarithm ln(1 + x), we must have |x| < 1 for convergence. Computing derivatives: f(x)=ln(1+x), f'(x)=1/(1+x), f''(x)=−1/(1+x)², f'''(x)=2/(1+x)³, and so on. At x=0: f(0)=0, f'(0)=1, f''(0)=−1, f'''(0)=2, f⁽⁴⁾(0)=−6. The general term for n ≥ 1 is (−1)ⁿ⁻¹ (n−1)! / n! = (−1)ⁿ⁻¹ / n. Thus:

对于自然对数 ln(1+x)，收敛需满足 |x| < 1。求导得：f'(x)=1/(1+x)，f''(x)=−1/(1+x)²，f'''(x)=2/(1+x)³，等等。在 x=0 处：f(0)=0, f'(0)=1, f''(0)=−1, f'''(0)=2, f⁽⁴⁾(0)=−6。当 n ≥ 1 时，通项为 (−1)ⁿ⁻¹/n。故得：

ln(1 + x) = x − x²/2 + x³/3 − x⁴/4 + … (for −1 < x ≤ 1)

ving the alternating harmonic series) but excludes x = −1 where the function is undefined.

注意收敛区间包含 x = 1（得到交错调和级数），但不含 x = −1，此时函数无定义。

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6. Binomial Series as a Maclaurin Expansion | 二项式级数作为麦克劳林展开

The function (1 + x)ᵐ, where m is any real number, expands via the generalised binomial theorem. Its Maclaurin series is:

函数 (1 + x)ᵐ（m 为任意实数）可通过广义二项式定理展开，麦克劳林级数为：

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

This expansion is valid for |x| < 1, unless m is a non‑negative integer (then it terminates and is valid for all x). You can derive it by differentiation: f(0)=1, f'(0)=m, f''(0)=m(m−1), etc.

该展开在 |x| < 1 时有效，除非 m 为非负整数（此时级数终止，且对所有 x 成立）。可通过求导得：f(0)=1, f'(0)=m, f''(0)=m(m−1) 等。

Common exam examples include √(1+x) = (1+x)^½ and 1/(1+x) = (1+x)⁻¹, which give 1 + ½ x − ⅛ x² + … and 1 − x + x² − x³ + … respectively.

常见考题如 √1+x = (1+x)^½ 展开为 1 + ½ x − ⅛ x² + …，而 1/(1+x) = (1+x)⁻¹ 展开为 1 − x + x² − x³ + …。

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7. Expanding Composite Functions | 复合函数的展开

You can often avoid repeated differentiation by substituting into known series. For instance, to expand e^2x, replace x with 2x in the eˣ series:

有时无需重复求导，直接将变量代入已知级数即可。例如展开 e^2x，只要把 eˣ 级数中的 x 换成 2x：

e^2x = 1 + (2x) + (2x)²/2! + (2x)³/3! + … = 1 + 2x + 2x² + (4/3)x³ + …

Similarly, sin(3x) = 3x − (3x)³/3! + (3x)⁵/5! − … = 3x − (9/2)x³ + (81/40)x⁵ − … Always simplify coefficients after substitution.

类似地，sin(3x) = 3x − (27x³)/6 + … 化简后为 3x − (9/2)x³ + (81/40)x⁵ − …。代入后务必化简系数。

This technique is extremely useful for functions like ln(1 − 2x): replace x with −2x in ln(1+x), yielding (−2x) − (−2x)²/2 + (−2x)³/3 − … = −2x − 2x² − (8/3)x³ − …

此法对 ln(1 − 2x) 同样有效：在 ln(1+x) 中将 x 换为 −2x，得 (−2x) − (−2x)²/2 + (−2x)³/3 − … = −2x − 2x² − (8/3)x³ − …。

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8. Using Maclaurin Series for Approximations | 用麦克劳林级数做近似计算

Truncating a Maclaurin series after a few terms gives a polynomial approximation. The error can be estimated using the next term (Lagrange remainder). In IGCSE, you are often asked to find an approximate value to a given degree of accuracy.

截断麦克劳林级数前几项便得到多项式近似。误差可由下一项（拉格朗日余项）估计。IGCSE 考试常要求按指定精度求近似值。

For example, approximate √4.1 using the expansion of (4 + x)^½ = 2(1 + x/4)^½. Let x = 0.1, expand (1 + u)^½ with u = 0.025, then multiply by 2: 2[1 + ½(0.025) − ⅛(0.025)² + …] ≈ 2[1 + 0.0125 − 0.000078] ≈ 2.0248. The true √4.1 ≈ 2.02485, excellent agreement.

例如，用 (4 + x)^½ = 2(1 + x/4)^½ 近似 √4.1。令 x = 0.1，展开 (1 + u)^½ 并取 u = 0.025，再乘 2：得 2[1 + ½(0.025) − ⅛(0.025)² + …] ≈ 2.0248，与真值 2.02485 非常吻合。

Always pay attention to the interval of convergence; using an x outside the valid range leads to a divergent (and useless) series.

务必注意收敛区间；若 x 超出有效范围，级数发散，毫无意义。

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9. Determining the Radius and Interval of Convergence | 确定收敛半径与区间

For an infinite Maclaurin series, convergence is not guaranteed for all x. The series for eˣ he series for ln(1+x) converges for −1 < x ≤ 1, while the binomial (1+x)ᵐ converges for |x| < 1 (unless m ∈ ℕ).

无穷麦克劳林级数并非对所有 x 都收敛。eˣ、sin x、cos x 的级数对所有实数收敛（半径 ∞）。ln(1+x) 的收敛区间为 −1 < x ≤ 1，二项式 (1+x)ᵐ 在 |x| < 1 时收敛（除非 m 为自然数）。

You can find the radius of convergence using the ratio test: ρ = lim |aₙ₊₁/aₙ| E exam, you are mainly expected to quote the known intervals and check endpoints if asked.

可用比值判别法求收敛半径：ρ = lim |aₙ₊₁/aₙ|，半径 R = 1/ρ。IGCSE 考试主要要求你记住已知区间，并根据题意检验端点。

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10. Common Exam Mistakes and Tips | 常见错误与解题技巧

Mistake 1: Forgetting to divide by factorials. The coefficient of xⁿ is f⁽ⁿ⁾(0)/n!, not just the derivative value.

错误一：忘记除以阶乘。xⁿ 的系数是 f⁽ⁿ⁾(0)/n!，而不仅仅是导数值。

Mistake 2: Using the wrong sign pattern. For sin x and cos x, signs alternate; for ln(1+x), the signs alternate starting with positive. Check the first few terms to be sure.

错误二：正负号弄错。sin x 和 cos x 是正负交替；ln(1+x) 从正号开始交替。验证前几项以保安全。

Mistake 3: Expanding about the wrong point. Remember Maclaurin demands x sks for an expansion about another point, you must use a Taylor series (not covered here).

错误三：在错误的点展开。记住麦克劳林级数要求 x=0；若题目要求在另一点展开，则需要泰勒级数（此处不涉及）。

Tip: When asked for the term in xⁿ of a composite function, it’s often faster to substitute into a known series than to differentiate n times.

技巧：求复合函数的 xⁿ 项时，代入已知级数通常比求n次导更快。

Tip: Always state the range of validity when giving a full expansion, unless the question says otherwise.

技巧：给出完整展开时，务必注明有效范围，除非题目另有说明。

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