📚 Maclaurin Series Expansion for CIE A-Level Mathematics | A-Level CIE 数学：麦克劳林展开 考点精讲
The Maclaurin series is a cornerstone of CIE A-Level Mathematics (9709), appearing in both Pure Mathematics 2 and 3. It gives you a systematic way to express many important functions as infinite polynomials, making approximations, limit calculations and further analysis much simpler. Mastering this topic means not only memorising the standard expansions but also knowing how to derive them, how to manipulate series for composite functions, and how to apply them under exam conditions.
麦克劳林展开是 CIE A-Level 数学 (9709) 的重点内容，在纯数 2 和纯数 3 中均有考查。它为我们提供了一种系统的方法，将许多重要的函数表示成无穷多项式，从而使近似计算、极限求解以及更深入的分析变得简单明了。真正掌握这一考点，不仅要熟记几组标准展开式，还需要清楚如何推导它们、如何处理复合函数的级数，以及如何在考试中灵活应用。
1. Definition of the Maclaurin Series | 麦克劳林级数的定义
A Maclaurin series is a Taylor series expansion of a function about 0. If a function f(x) can be differentiated infinitely many times at x = 0, its Maclaurin series is given by the formula below. The series is a power series in x, and the coefficients are built from the derivatives of f at zero.
麦克劳林级数是函数在 x = 0 处的泰勒级数展开。如果函数 f(x) 在 x = 0 处可以无穷次求导，它的麦克劳林级数就由下面的公式给出。这是一个关于 x 的幂级数，每一项的系数由 f 在零点的各阶导数构成。
f(x) = f(0) + f'(0) x + f”(0) x²/2! + f”'(0) x³/3! + … + f⁽ⁿ⁾(0) xⁿ/n! + …
f(x) = Σₙ₌₀∞ (f⁽ⁿ⁾(0) / n!) xⁿ
You must be able to write down the first few terms of this expansion for a given function by computing the derivatives at 0 one by one. The factorial denominators come from the repeated differentiation of xⁿ.
对于给定的函数，你需要能够通过逐次计算在 0 处的导数值，写出展开式的前几项。分母中的阶乘来源于对 xⁿ 反复求导的过程。
2. The Standard Maclaurin Expansions to Memorise | 必须熟记的标准麦克劳林展开式
CIE expects you to know the following five standard expansions by heart. They serve as building blocks for more complicated problems. Always note the range of validity (the domain of convergence) because the series only represents the function when it converges.
CIE 考纲要求熟记以下五个标准展开式，它们是解决更复杂问题的基础。务必注意每个级数的有效范围（收敛区间），因为只有在收敛时级数才真正等于函数。
- eˣ = 1 + x + x²/2! + x³/3! + … + xⁿ/n! + … (valid for all real x)
- sin x = x – x³/3! + x⁵/5! – x⁷/7! + … (valid for all real x)
- cos x = 1 – x²/2! + x⁴/4! – x⁶/6! + … (valid for all real x)
- 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! + … (valid for |x| < 1, and for all x if n is a non‑negative integer)
Notice that sin x contains only odd powers, cos x contains only even powers, and ln(1+x) has no factorial denominators. The binomial expansion simplifies to a finite polynomial when n is a positive integer.
注意 sin x 只含奇次幂，cos x 只含偶次幂，ln(1+x) 的分母没有阶乘。二项式展开当 n 为正整数时会退化为有限次多项式。
3. Deriving Maclaurin Series Step by Step | 麦克劳林展开的推导步骤
To find the Maclaurin series of a function like f(x) = √(1+2x), follow these four steps: (1) evaluate f(0); (2) find f'(x), f”(x), f”'(x) as needed and evaluate each derivative at 0; (3) construct the coefficients using f⁽ⁿ⁾(0)/n!; (4) assemble the series up to the required power, usually x³ or x⁴ in CIE exams.
要找出像 f(x) = √(1+2x) 这样的函数的麦克劳林级数，可遵循四个步骤：(1) 计算 f(0)；(2) 根据需求求出 f'(x), f”(x), f”'(x) 等，并计算每阶导数在 0 处的值；(3) 用 f⁽ⁿ⁾(0)/n! 构建系数；(4) 合并级数，通常写到 x³ 或 x⁴ 项，这也是 CIE 考试的常见要求。
For example, with f(x) = (1 + 2x)^½, you get f(0)=1, f'(0)=1, f”(0)=-1, f”'(0)=3, giving the expansion 1 + x – ½ x² + ½ x³ + … . Always simplify coefficients completely.
例如，对于 f(x) = (1 + 2x)^½，可得 f(0)=1, f'(0)=1, f”(0)=-1, f”'(0)=3，从而写出展开式 1 + x – ½ x² + ½ x³ + … 。系数一定要化简到最简形式。
4. Substitution in Standard Series | 标准级数的代换法
Once you know the standard series, you can obtain new expansions by substituting a multiple of x, or a power of x, or a simple expression like 2x, -x, x² into the basic formulas. This is often much faster than differentiating from first principles.
掌握了标准级数之后，你可以通过将 x 的某个倍数、幂次或简单的表达式（如 2x, -x, x²）代入基本公式来获得新函数的展开式。这通常比从定义出发逐阶求导快得多。
For instance, to expand e²ˣ, replace x by 2x in eˣ: e²ˣ = 1 + (2x) + (2x)²/2! + (2x)³/3! + … = 1 + 2x + 2x² + (4/3)x³ + … . Similarly, sin(3x) = 3x – (3x)³/3! + … = 3x – (9/2)x³ + … .
例如，展开 e²ˣ 时，用 2x 替换 eˣ 级数中的 x：e²ˣ = 1 + (2x) + (2x)²/2! + (2x)³/3! + … = 1 + 2x + 2x² + (4/3)x³ + … 。类似地，sin(3x) = 3x – (3x)³/3! + … = 3x – (9/2)x³ + … 。
Always make sure the substitution respects the convergence interval. For ln(1 + 3x), the expansion is 3x – (3x)²/2 + (3x)³/3 – … and it is valid only for -⅓ < x ≤ ⅓.
代换时一定要注意收敛区间。对于 ln(1 + 3x)，展开式为 3x – (3x)²/2 + (3x)³/3 – …，它只有在 -⅓ < x ≤ ⅓ 时才有效。
5. Multiplying and Dividing Series | 级数的乘法与除法
When a function is a product of two functions whose expansions you know, such as eˣ cos x, you can multiply the series term by term, keeping only the powers you need. CIE often asks for expansions up to the term in x³, so cross‑multiply and discard terms of degree 4 and higher as you work.
当一个函数是两个已知展开式的函数的乘积时，例如 eˣ cos x，你可以将两个级数逐项相乘，只保留所需的幂次。CIE 考试通常要求展开到 x³ 项，因此在相乘过程中可以直接舍弃四次及更高次的项。
Write eˣ ≈ 1 + x + x²/2 + x³/6, cos x ≈ 1 – x²/2, then multiply: (1 + x + ½x² + ⅙x³)(1 – ½x²) = 1 + x + ½x² + ⅙x³ – ½x² – ½x³ + … = 1 + x – ⅓x³ + … (neglecting higher terms). This is much quicker than differentiating the product.
写出 eˣ ≈ 1 + x + x²/2 + x³/6, cos x ≈ 1 – x²/2，然后相乘：(1 + x + ½x² + ⅙x³)(1 – ½x²) = 1 + x + ½x² + ⅙x³ – ½x² – ½x³ + … = 1 + x – ⅓x³ + …（忽略高次项）。这比直接对乘积求导快得多。
Division can be handled by writing the quotient as a product with the reciprocal series, or by using the binomial expansion when the denominator is of the form (1 + u)⁻¹.
除法可以通过将商写为与倒数级数的乘积来处理，或者当分母是 (1 + u)⁻¹ 的形式时直接使用二项式展开。
6. Approximating Functions and Finding Numerical Values | 用级数求近似值与数值估计
One of the most practical applications of Maclaurin series is finding approximations to function values for small x. By truncating the series, you obtain a polynomial that closely matches the function near zero. The more terms you retain, the better the approximation.
麦克劳林级数最实际的应用之一是当 x 很小时，估算函数的值。截断级数后，你会得到一个在零点附近与函数高度吻合的多项式。保留的项数越多，近似效果越好。
For example, using sin x ≈ x – x³/6, you can approximate sin(0.1) as 0.1 – (0.1)³/6 ≈ 0.099833. Compare this with the true value to see how accurate the approximation is. CIE may ask you to estimate an integral or a compound expression by substituting a small value into a series.
例如，使用 sin x ≈ x – x³/6，可以将 sin(0.1) 近似为 0.1 – (0.1)³/6 ≈ 0.099833。与真实值对比即可看出近似的精确程度。CIE 有时会要求将一个小数值代入级数来估算某个积分或复合表达式的值。
7. Evaluating Limits Using Maclaurin Series | 利用麦克劳林级数求极限
When a limit takes an indeterminate form like 0/0, replacing the numerator and denominator with their Maclaurin expansions often resolves the limit cleanly. This technique is especially powerful when L’Hôpital’s rule would be tedious to apply repeatedly.
当某个极限呈现 0/0 型未定式时，将分子和分母分别用它们的麦克劳林级数替换，常常能清晰地解出极限。当反复使用洛必达法则过于繁琐时，这一技巧尤为有效。
For instance, limₓ→₀ (sin x – x)/x³ can be evaluated by substituting sin x = x – x³/6 + x⁵/120 – … . Then numerator becomes (x – x³/6 + …) – x = -x³/6 + … . Dividing by x³ gives -1/6 as x→0. This is a classic CIE problem.
例如，求极限 limₓ→₀ (sin x – x)/x³，代入 sin x = x – x³/6 + x⁵/120 – …，分子变为 (x – x³/6 + …) – x = -x³/6 + …，除以 x³ 后得到 -1/6。这是 CIE 考试中的经典题型。
8. Finding Derivatives Using Series | 利用级数求导数值
Because the Maclaurin series is unique, the coefficient of xⁿ in the expansion of f(x) must be exactly f⁽ⁿ⁾(0)/n!. This allows you to read off high‑order derivatives at 0 without actually differentiating that many times, as long as you can obtain the series by substitution or simple manipulation.
由于麦克劳林级数是唯一的，f(x) 的展开式中 xⁿ 项的系数必定等于 f⁽ⁿ⁾(0)/n!。因此，只要你能通过代换或简单变形得到级数，就可以直接读出在 0 处的高阶导数值，而不必真的求导多次。
For f(x) = e^(3x²), substitute u = 3x² into eᵘ: f(x) = 1 + 3x² + (9/2)x⁴ + … . The coefficient of x⁴ is 9/2, so f⁽⁴⁾(0)/4! = 9/2 ⇒ f⁽⁴⁾(0) = 108. This saves enormous time in the exam.
对于 f(x) = e^(3x²)，将 u = 3x² 代入 eᵘ 的展开式：f(x) = 1 + 3x² + (9/2)x⁴ + … 。x⁴ 的系数是 9/2，因此 f⁽⁴⁾(0)/4! = 9/2 ⇒ f⁽⁴⁾(0) = 108。这能在考试中节省大量时间。
9. Common Pitfalls and How to Avoid Them | 常见错误与避坑指南
Many CIE candidates lose marks not because they cannot do the expansion, but because of small slips. The most frequent mistakes include forgetting factorial denominators, mishandling signs in sin and cos, miswriting the binomial coefficient for (1+x)ⁿ, and ignoring the validity range. Also, when using substitution, failing to adjust powers correctly (e.g., replacing x by x² squares the powers) causes errors.
不少 CIE 考生丢分不是因为不会展开，而是由于小失误。最常见的错误包括：忘记阶乘分母、弄错 sin 和 cos 的正负号、写错 (1+x)ⁿ 的二项式系数，以及忽略有效范围。此外，使用代换法时没有正确调整幂次（例如用 x² 替换 x 后未能准确算出新的幂次）也会导致出错。
Always write the first three or four terms carefully, simplify each coefficient fully, and double‑check the sign pattern. If the question asks for an expansion ‘up to and including the term in x³’, never include x⁴ or omit the constant term. Also, remember that a Maclaurin series is a sum of terms; losing a term early on propagates through the whole working.
一定要仔细写出前三或四项，将每个系数彻底化简，并反复检查正负规律。如果题目要求“展开到含 x³ 项为止”，就不要保留 x⁴ 项，也不要丢掉常数项。还要记住，麦克劳林级数是一系列项的和，早期丢失一项会导致整题出错。
10. Typical Exam-Style Question with Full Worked Solution | 考试典型题与完整解题示范
Consider this CIE-style question: Find the Maclaurin series for f(x) = ln(1 + sin x) up to and including the term in x³.
来看一道 CIE 风格的题目：求 f(x) = ln(1 + sin x) 的麦克劳林展开式，写至含 x³ 项。
Step 1: Write the standard series for sin x and ln(1+u). sin x ≈ x – x³/6, ln(1+u) ≈ u – u²/2 + u³/3. Step 2: Let u = sin x ≈ x – x³/6. Step 3: Compute u² = (x – x³/6)² ≈ x² – x⁴/3, ignore x⁴ and above. u³ ≈ x³. Step 4: Substitute into ln(1+u): f(x) ≈ (x – x³/6) – ½(x²) + ⅓(x³) = x – x²/2 + ( -1/6 + 1/3 )x³ = x – x²/2 + x³/6. This method is rapid and reliable.
第一步：写出 sin x 和 ln(1+u) 的标准级数。sin x ≈ x – x³/6, ln(1+u) ≈ u – u²/2 + u³/3。第二步：令 u = sin x ≈ x – x³/6。第三步：计算 u² = (x – x³/6)² ≈ x² – x⁴/3，忽略 x⁴ 及以上项；u³ ≈ x³。第四步：代入 ln(1+u)：f(x) ≈ (x – x³/6) – ½(x²) + ⅓(x³) = x – x²/2 + ( -1/6 + 1/3 )x³ = x – x²/2 + x³/6。这种方法又快又准。
If you were asked to determine f”'(0) from this expansion, you would simply read the x³ coefficient: 1/6 = f”'(0)/6 ⇒ f”'(0) = 1. Master this workflow, and you can tackle virtually any Maclaurin question with confidence.
如果题目要求从这个展开式中求出 f”'(0)，只需读出 x³ 的系数：1/6 = f”'(0)/6 ⇒ f”'(0) = 1。掌握这一套流程，你就能自信地应对几乎任何麦克劳林展开题。
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