📚 IGCSE Edexcel Maths: Maclaurin Expansion Key Points | IGCSE Edexcel 数学：麦克劳林展开 考点精讲
The Maclaurin expansion is a powerful tool for expressing a function as an infinite sum of terms calculated from its derivatives at zero. Although it is formally part of A Level Further Mathematics, ambitious IGCSE Edexcel students who are comfortable with differentiation and binomial expansions can gain valuable insight by exploring this concept early. This article breaks down the key points, common expansions, and exam-style skills you need to understand Maclaurin series.
麦克劳林展开是将函数表示为基于其在零点处的导数计算出的无穷多项之和的强大工具。虽然它正式属于 A Level 进阶数学的范畴，但已熟悉微分和二项式展开的、有抱负的 IGCSE Edexcel 学生可以提前探索这一概念并从中获益。本文将精讲麦克劳林级数的关键考点、常见展开式以及你需要掌握的应试技巧。
1. What is Maclaurin Expansion? | 什么是麦克劳林展开？
A Maclaurin series is a Taylor series expansion of a function about 0. It allows us to write many differentiable functions as a sum of powers of x, where the coefficients are determined by the function’s derivatives evaluated at x = 0. If the function is infinitely differentiable and the series converges, the expansion is exact within its interval of convergence.
麦克劳林级数是函数在 0 处的泰勒级数展开。它让我们能够将许多可微函数写成 x 的幂次之和，其中各项的系数由函数在 x = 0 处的导数值决定。如果函数无穷可微且级数收敛，该展开式在其收敛区间内是精确的。
The basic idea is to match the function value and all its derivatives at a single point. For IGCSE students, think of it as an extension of the idea that a polynomial can be built up from its derivatives at a point – but now extended to non-polynomial functions like eˣ, sin x or ln(1+x).
基本思想是在同一点匹配函数值和它的所有导数值。对于 IGCSE 学生来说，可以将其看作是“多项式可以由它在某点的导数构建”这一想法的延伸——但现在延伸到了像 eˣ、sin x 或 ln(1+x) 这样的非多项式函数。
2. The General Formula | 通项公式
The Maclaurin series for a function f(x) is given by the infinite sum:
函数 f(x) 的麦克劳林级数由以下无穷和式给出：
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 of f evaluated at x = 0. The factorial n! grows very quickly, which often helps the series converge for small values of x. You must be comfortable taking repeated derivatives to find the coefficients.
这里 f⁽ⁿ⁾(0) 表示 f 在 x=0 处的 n 阶导数。阶乘 n! 增长速度非常快，这通常有助于级数在 x 取值较小时收敛。你必须能熟练地求多次导数以找到各项系数。
In practice, you may only need the first few terms (up to x², x³ or x⁴) to approximate a function near zero. Each term adds a closer fit around the origin.
实践中你可能只需要前几项（直到 x²、x³ 或 x⁴）来近似函数在零点附近的行为。每一项的加入都让多项式在原点附近更好地贴合原函数。
3. Expansions of Basic Functions | 基本函数的展开式
Three fundamental Maclaurin expansions are essential to memorise. They are derived by differentiating repeatedly and evaluating at zero:
三个基础的麦克劳林展开式必须牢记。它们可以通过反复求导并在零点取值得到：
| Function | Maclaurin Series |
|---|---|
| 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! + … |
Notice the alternating signs in the sine and cosine series, and that sine contains only odd powers while cosine contains only even powers. The exponential series has all positive signs.
注意正弦和余弦级数中的交替符号，以及正弦级数只包含奇次幂而余弦级数只包含偶次幂。指数级数所有项均为正号。
These three series are the building blocks for many more complicated expansions. By combining them, differentiating, or integrating term by term, you can obtain series for e²ˣ, sin(x²), or eˣ cos x without starting from scratch.
这三个级数是众多更复杂展开式的基石。通过组合它们、逐项微分或积分，你可以获得 e²ˣ、sin(x²) 或 eˣ cos x 的级数而无需从头推导。
4. Connection to Binomial Expansion | 与二项式展开的联系
The binomial expansion you learned at IGCSE is a special case of the Maclaurin series. For f(x) = (1+x)ⁿ, where n can be any real number, repeated differentiation gives f⁽ᵏ⁾(0) = n(n-1)…(n-k+1). The Maclaurin series becomes:
你在 IGCSE 学到的二项式展开是麦克劳林级数的一个特例。对于 f(x) = (1+x)ⁿ，其中 n 可为任意实数，反复求导可得 f⁽ᵏ⁾(0) = n(n-1)…(n-k+1)。麦克劳林级数成为：
(1+x)ⁿ = 1 + nx + n(n-1)x²/2! + n(n-1)(n-2)x³/3! + …
When n is a positive integer, the series terminates after (n+1) terms, giving the exact polynomial expansion. When n is not a positive integer, the series is infinite and only converges for |x| < 1. This explains why the IGCSE binomial expansion formula works and how it extends to negative and fractional powers.
当 n 为正整数时，该级数在第 (n+1) 项后终止，产生精确的多项式展开。当 n 不是正整数时，级数为无穷级数且仅在 |x| < 1 时收敛。这解释了 IGCSE 二项式展开公式为何有效，以及它如何扩展到负指数和分数指数。
For example, to approximate √(1.02), set f(x) = (1+x)^½, n=1/2, and use the expansion up to x²: √(1+x) ≈ 1 + x/2 – x²/8. Substituting x=0.02 gives a quickly accurate estimate.
例如，近似计算 √(1.02)，令 f(x) = (1+x)^½，n=1/2，使用展开到 x²：√(1+x) ≈ 1 + x/2 – x²/8。代入 x=0.02 即可快速得到精确估计值。
5. Maclaurin Series for ln(1+x) | ln(1+x) 的麦克劳林级数
The natural logarithm function ln(1+x) is another important expansion. Its derivatives at zero are f(0)=0, f'(0)=1, f”(0)=-1, f”'(0)=2!, f⁽⁴⁾(0)=-3!, etc. The resulting series is:
自然对数函数 ln(1+x) 是另一个重要的展开式。它在零点处的导数为 f(0)=0、f'(0)=1、f”(0)=-1、f”'(0)=2!、f⁽⁴⁾(0)=-3! 等等。得出的级数为：
ln(1+x) = x – x²/2 + x³/3 – x⁴/4 + x⁵/5 – …
This series is valid for -1 < x ≤ 1. Note the alternating signs and the fact that the coefficient of xⁿ is (-1)ⁿ⁻¹/n for n≥1. It converges much more slowly than the series for eˣ, so more terms are needed for a good approximation away from zero.
该级数在 -1 < x ≤ 1 内有效。注意交替的符号以及 xⁿ 的系数为 (-1)ⁿ⁻¹/n（n≥1）。它比 eˣ 的级数收敛慢得多，因此在远离零的地方需要更多项才能达到良好近似。
Understanding ln(1+x) helps with expansions of related functions such as ln(1-x) (just replace x by -x) or ln[(1+x)/(1-x)] by subtracting series.
理解 ln(1+x) 有助于处理相关函数的展开，如 ln(1-x)（只需将 x 替换为 -x）或通过级数相减得到 ln[(1+x)/(1-x)] 的展开。
6. Using Expansions for Approximations | 使用展开进行近似计算
One of the most practical applications of Maclaurin series is approximating function values. By truncating the series after a few terms, we obtain a polynomial that closely mimics the function for small x. The error can be estimated using the next term or Lagrange remainder.
麦克劳林级数最实际的应用之一就是近似计算函数值。通过在几项之后截断级数，我们得到一个在 x 较小时非常接近原函数的多项式。其误差可用下一项或拉格朗日余项来估计。
For example, to estimate e⁰·¹, use the expansion eˣ ≈ 1 + x + x²/2 + x³/6. With x=0.1: 1 + 0.1 + 0.005 + 0.0001667 = 1.1051667. The true value is about 1.1051709, so the approximation is excellent. This demonstrates why a few derivatives at zero can characterise the function so well near the origin.
例如，估算 e⁰·¹，使用展开式 eˣ ≈ 1 + x + x²/2 + x³/6。代入 x=0.1 得到 1 + 0.1 + 0.005 + 0.0001667 = 1.1051667。真实值约为 1.1051709，可见近似效果极佳。这说明了为何在原点附近的几个导数值就能很好地刻画函数。
For the IGCSE extension level, you might be asked to use a given Maclaurin series to find an approximate value or to compare it with a calculator value. Always note the order of the approximation (up to x², x³, etc.).
在 IGCSE 拓展层面，你可能需要使用给定的麦克劳林级数求近似值，或与计算器结果比较。务必注意近似的阶数（精确到 x²、x³ 等）。
7. Error and Interval of Convergence | 误差与收敛区间
Not all Maclaurin series converge for all x, and the truncated series carries an error. The interval of convergence is the set of x for which the infinite series sums to the function. For eˣ, sin x and cos x, the interval is all real numbers (–∞, ∞). For (1+x)ⁿ and ln(1+x), convergence is typically restricted to |x| < 1.
并非所有麦克劳林级数都对所有 x 收敛，截断级数也带有误差。收敛区间是使无穷级数求和等于原函数的 x 的集合。对于 eˣ、sin x 和 cos x，收敛区间为全体实数 (–∞, ∞)。对于 (1+x)ⁿ 和 ln(1+x)，收敛通常局限于 |x| < 1。
A simple way to think about error: when you stop at the term in xⁿ, the error is roughly bounded by the magnitude of the next term (if the series is alternating) or by a Lagrangian remainder formula. For exams, you may just need to know that the approximation improves as x gets smaller and as more terms are included.
关于误差的简单理解：当你停在 xⁿ 项时，误差大致由下一项的大小控制（若级数为交错级数），或由拉格朗日余项公式界定。在考试中，你可能只需知道当 x 越小、包含的项越多时近似效果越好即可。
8. Multiplying and Composing Series | 级数的乘法与复合
You can obtain Maclaurin series for products and compositions without directly differentiating multiple times. For instance, to find the expansion of eˣ sin x up to x³, you can multiply the series:
你可以无需多次直接求导而获得乘积和复合函数的麦克劳林级数。例如，要找到 eˣ sin x 展开到 x³ 的项，可以将级数相乘：
eˣ = 1 + x + x²/2 + x³/6 + …
sin x = x – x³/6 + …
Multiplying and collecting terms up to x³: (1)(x) + (x)(x) gives an x² term? Actually careful: (1 + x + x²/2 + x³/6)(x – x³/6) = x + x² + (x²/2)x? Let us compute: (1·x) = x, (x·x) = x², (1·(–x³/6)) = –x³/6, (x²/2 · x) = x³/2, (x·x²) but x·x² gives x³ from x and x² in the expansions? Wait we must only consider up to x³. So eˣ up to x³: 1+x+x²/2+x³/6. sin x up to x³: x – x³/6. Product: 1*(x – x³/6) = x – x³/6; x*(x – x³/6) = x² – x⁴/6 (ignore x⁴); x²/2*(x – x³/6) = x³/2 – x⁵/12 (keep x³/2); x³/6*(x – x³/6) ≈ x⁴/6 (ignore). Summing: x + x² + (–x³/6 + x³/2) = x + x² + (x³/3). So eˣ sin x ≈ x + x² + x³/3.
相乘并收集到 x³ 项：(1 + x + x²/2 + x³/6)(x – x³/6) 计算得 x + x² + ( –x³/6 + x³/2 ) = x + x² + x³/3。所以 eˣ sin x ≈ x + x² + x³/3。
This method is much faster than finding the third derivative of eˣ sin x at zero. Similarly, composing functions, such as e^(sin x) or √(1+x²), can be handled by substituting one series into another.
这比在零点求 eˣ sin x 的三阶导数要快得多。类似地，复合函数如 e^(sin x) 或 √(1+x²) 可通过将一个级数代入另一个进行处理。
9. Step-by-Step Example | 分步例题
Question: Find the Maclaurin series for f(x) = 1/(1–x) up to the term in x³. Hence, find the series for 1/(1+x) up to x³.
问题：求 f(x) = 1/(1–x) 的麦克劳林展开到 x³ 项。由此求出 1/(1+x) 展开到 x³ 的级数。
Step 1: Compute derivatives at 0. f(0)=1. f'(x) = 1/(1–x)² → f'(0)=1. f”(x) = 2/(1–x)³ → f”(0)=2. f”'(x) = 6/(1–x)⁴ → f”'(0)=6.
步骤 1：计算在 0 处的导数。f(0)=1。f'(x)=1/(1–x)² → f'(0)=1。f”(x)=2/(1–x)³ → f”(0)=2。f”'(x)=6/(1–x)⁴ → f”'(0)=6。
Step 2: Plug into Maclaurin formula: f(x) = 1 + 1·x + 2·x²/2! + 6·x³/3! + … = 1 + x + x² + x³ + ….
步骤 2：代入麦克劳林公式：f(x) = 1 + 1·x + 2·x²/2! + 6·x³/3! + … = 1 + x + x² + x³ + …。
Step 3: For 1/(1+x), replace x by –x in the series: 1/(1–(–x)) = 1 + (–x) + (–x)² + (–x)³ + … = 1 – x + x² – x³ + ….
步骤 3：对于 1/(1+x)，将级数中的 x 替换为 –x：1/(1–(–x)) = 1 + (–x) + (–x)² + (–x)³ + … = 1 – x + x² – x³ + …。
This example shows both direct derivation and the power of substitution, which is a favourite exam technique.
这个例子展示了直接推导和代入法的威力，这是一项很受考试青睐的技巧。
10. Exam Tips and Summary | 考试技巧与总结
When approaching Maclaurin expansion questions, remember: (1) Memorise the key series for eˣ, sin x, cos x, (1+x)ⁿ, and ln(1+x). (2) Practise taking higher derivatives carefully, as a single sign error will throw off the whole series. (3) Always state the first few terms clearly up to the required power. (4) Use substitution and combination of known series to save time. (5) If a series is given, you can differentiate or integrate term by term to find related series (e.g., differentiate ln(1+x) to get 1/(1+x)).
解答麦克劳林展开题目的要点： (1) 牢记 eˣ、sin x、cos x、(1+x)ⁿ 和 ln(1+x) 的关键级数。 (2) 仔细练习求高阶导数，一个符号出错就会让整个级数失效。 (3) 务必按要求列出前几项，直到指定的幂次。 (4) 使用代入法和已知级数的组合来节约时间。 (5) 若给定一个级数，可以逐项微分或积分来求得相关级数（例如，微分 ln(1+x) 得到 1/(1+x)）。
Even though Maclaurin expansion goes beyond the standard IGCSE syllabus, understanding its logic strengthens your insight into functions, derivatives, and the idea that smooth functions can be ‘built’ from their behaviour at a single point. This foundation will serve you well in further mathematics and science subjects.
尽管麦克劳林展开超出了标准 IGCSE 大纲，但理解其
Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply