📚 A-Level OCR Maths: Maclaurin Series Essentials | A-Level OCR 数学：麦克劳林展开考点精讲
The Maclaurin series is a powerful tool in A-Level OCR Mathematics that allows us to express complicated functions as infinite polynomials. In your exam, you’ll be expected to derive series from first principles, recall standard expansions, use them for approximations, and even solve differential equations. This article breaks down every key topic into bite-sized, exam-focused sections, pairing clear English explanations with their Chinese counterparts to deepen bilingual understanding.
在 A-Level OCR 数学中，麦克劳林级数是一种将复杂函数表达为无穷多项式的重要工具。考试中你既要从第一原理推导级数，也要熟记标准展开式，并能用于近似计算乃至解微分方程。本文把每个关键考点拆解成中英对照的精讲小节，帮助双语学习者透彻掌握。
1. Understanding Maclaurin Series – Definition and Formula | 理解麦克劳林级数 – 定义与公式
A Maclaurin series is a Taylor series expansion of a function f(x) about x = 0. It says that if f is infinitely differentiable at 0, then f(x) can be written as f(0) + f'(0)x + f”(0)x²/2! + f”'(0)x³/3! + … + f⁽ⁿ⁾(0)xⁿ/n! + … This formula is the starting point for every exam question on the topic. Note that the coefficient of xⁿ is always the nth derivative at 0 divided by n!.
麦克劳林级数就是函数 f(x) 在 x=0 处的泰勒展开。只要 f 在 0 处无限可导，f(x) 就可以写成 f(0)+f'(0)x+f”(0)x²/2!+f”'(0)x³/3!+…+f⁽ⁿ⁾(0)xⁿ/n!+…。考试中所有问题都从这个公式出发。注意 xⁿ 项系数总是第 n 阶导数值除以 n!。
For OCR, you are expected to know the general term and be able to compute derivatives up to fourth order explicitly. Often the question will say ‘find the Maclaurin series up to and including the term in x³’. That means you need f(0), f'(0), f”(0) and f”'(0).
OCR 考试要求你掌握通项公式，并能显式求出至少前四阶导数。题目常会写“求麦克劳林级数，含 x³ 项为止”，即需要求出 f(0)、f'(0)、f”(0) 和 f”'(0)。
2. Standard Maclaurin Series Every Student Must Know | 考生必须掌握的标准麦克劳林展开式
The exam board expects you to memorise five fundamental expansions. These are tested directly and used as building blocks for substitution questions.
考试委员会要求你熟记五个基本展开式，它们会直接考查，这也是代换法的基础。
| 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 (if n not a positive integer) |
Memorising these will save you precious time. The exponential, sine and cosine series converge for all real x, but the logarithmic and binomial series have restricted validity intervals – always check the range in approximation questions.
熟记这五个可以节省大量考试时间。指数、正弦和余弦级数对所有实数 x 都收敛，而对数和二项式级数则存在收敛区间限制，在近似计算中务必检查 x 的范围。
3. Validity and Radius of Convergence | 有效性及收敛半径
Every power series only equals the original function within a certain interval called the domain of convergence. For OCR, you must state the validity after writing a Maclaurin series. For instance, the series for ln(1+x) is valid for −1 < x ≤ 1, and (1+x)ⁿ for |x| < 1 when n is a fraction or negative.
每一个幂级数只有在特定的收敛区间内才等于原函数。OCR 要求你在写出麦克劳林级数后必须注明有效性。例如 ln(1+x) 的展开对 −1 < x ≤ 1 有效，而 (1+x)ⁿ 当 n 为分数或负数时需满足 |x| < 1。
In exam problems, when you substitute, say, 2x into a known series, the validity transforms accordingly: if (1+u)ⁿ is valid for |u|<1, then (1+2x)ⁿ is valid for |2x|<1, i.e. |x|<0.5. Forgetting to adjust the validity interval is a common error that loses marks.
考试中如果对已知级数进行代换，比如将 2x 代入，有效区间也要同步调整：若 (1+u)ⁿ 对 |u|<1 有效，则 (1+2x)ⁿ 对 |2x|<1，即 |x|<0.5 才有效。忘记调整有效性标记是常见的失分点。
4. Deriving a Maclaurin Series Step by Step | 逐步推导麦克劳林展开式
The basic method is to differentiate repeatedly, evaluate at 0, and plug into the formula. Let’s illustrate with f(x) = sec x. We need f(0)=1, f'(0)=0, f”(0)=1, f”'(0)=0. Thus sec x ≈ 1 + x²/2! + … (the coefficient of x³ is zero). This shows that not every degree term appears – the series can have only even powers for an even function.
基本方法就是反复求导、计算 x=0 处的值并代入公式。以 f(x) = sec x 为例，f(0)=1, f'(0)=0, f”(0)=1, f”'(0)=0。因此 sec x ≈ 1 + x²/2! + … （x³ 系数为零）。这表明并非每一次幂项都存在——偶函数的级数可能只含偶次幂。
OCR questions often provide repetitive differentiation tables: derivative, value at 0, and coefficient. Organising your work in a table reduces slip-ups. Also, note that you may be asked to ‘find the first three non-zero terms’, so do not stop at x³ if the x³ term is zero; continue until you have three meaningful terms.
OCR 考题常会提供重复求导表格：导数、x=0 处的值和系数。用表格整理可避免粗心错误。另外，题目可能要求“写出前三个非零项”，所以若 x³ 项为零，不要停下来；要继续加到获得三个有效项为止。
5. Using Substitutions to Generate New Series | 利用代换法生成新的级数
Once you know the standard series, you can create expansions for related functions quickly. For example, to expand e−x², simply replace x with −x² in the series for eˣ: 1 + (−x²) + (−x²)²/2! + … = 1 − x² + x⁴/2! − … . This works because the exponential series converges for all real numbers, so the substitution is valid everywhere.
一旦熟记标准级数，就能迅速生成相关函数的展开式。比如要展开 e−x²，只需将 eˣ 级数中的 x 替换为 −x² 即可：1+(−x²)+(−x²)²/2!+… = 1−x²+x⁴/2!−…。因为指数级数对所有实数收敛，所以代换后处处有效。
Another common substitution is to obtain sin 2x or cos(x/2). Always check validity: if the original series was valid for all x, the substituted series is also valid for all x. But if you substitute into ln(1+x), ensure the new input lies in (−1,1].
另一个常见代换是求 sin 2x 或 cos(x/2) 的展开。务必检查有效性：若原始级数对所有 x 有效，代换后依然处处有效；但若代入 ln(1+x) 中，则须保证新的变量取值落在 (−1,1] 内。
6. Maclaurin Series for Composite Functions | 复合函数的麦克劳林级数
OCR exams occasionally ask for the expansion of a composite like esin x or ln(cos x). There are two approaches: fruit-by-fruit differentiation, or building up from known series. The latter is more elegant. For esin x, you can write: sin x = x − x³/6 + …; then eu = 1 + u + u²/2! + u³/3! + …; substitute u = x − x³/6 + … and carefully collect terms up to the required power.
OCR 卷子偶尔会要求展开复合函数，如 esin x 或 ln(cos x)。解决方法有二：逐次求导，或基于已知级数组合。后者更简洁。例如对于 esin x，可写 sin x = x − x³/6 + …；然后 eu = 1+u+u²/2!+u³/3!+…；代入 u = x − x³/6 + … 并仔细收集所需次幂的项。
After substitution, multiply out and discard terms beyond your target degree. Watch out for terms generated from the square or cube of u – they contribute to lower-degree coefficients. A systematic grid or column method helps avoid missing cross terms.
代换之后，展开相乘并舍弃超出目标次数的项。注意 u 的平方或立方会生成低次幂的贡献。使用系统性的网格或列式方法可避免遗漏交叉项。
7. Approximating Functions and Calculating Errors | 函数近似计算与误差估计
A classic OCR requirement is to use a Maclaurin polynomial to estimate a value, like e0.2 or sin 0.3, and then bound the error using the Lagrange remainder. The Lagrange form states that the error after n terms is Rn(x) = f⁽ⁿ⁺¹⁾(c) xⁿ⁺¹/(n+1)! for some c between 0 and x. You often bound |f⁽ⁿ⁺¹⁾(c)| by its maximum modulus on [0,x].
OCR 非常经典的要求是用麦克劳林多项式估算如 e0.2 或 sin 0.3 的值，再通过拉格朗日余项控制误差。拉格朗日余项形式为 Rn(x) = f⁽ⁿ⁺¹⁾(c) xⁿ⁺¹/(n+1)!，其中 c 介于 0 与 x 之间。通常需用 |f⁽ⁿ⁺¹⁾(c)| 在 [0,x] 上的最大值来放缩。
For example, to estimate √e with the first three terms of eˣ at x=0.5, use 1+0.5+0.5²/2=1.625. The next term in the series is 0.5³/6 ≈ 0.0208. If the fourth derivative eˣ is bounded by e0.5 < 2, then the error is less than 2×0.5⁴/24 = 0.0052. So the true value lies between 1.625 and 1.6302.
举例来说，用 eˣ 前三项在 x=0.5 处估算 √e：1+0.5+0.5²/2=1.625。下一项是 0.5³/6 ≈ 0.0208。若四阶导数 eˣ 在区间上的界为 e0.5 < 2，则误差小于 2×0.5⁴/24=0.0052。因此真实值介于 1.625 与 1.6302 之间。
8. Linking Maclaurin Series to Differential Equations | 麦克劳林级数与微分方程的联系
Newer OCR papers have introduced problems where you find a Maclaurin series solution to a differential equation. Given dy/dx = f(x,y) and an initial condition y(0) = a, you can differentiate the equation repeatedly to obtain y'(0), y”(0), y”'(0) etc., and then construct the series. This is a neat application that bypasses the need to solve the DE analytically.
较新的 OCR 考卷引入了利用麦克劳林级数解微分方程的题型。给定 dy/dx = f(x,y) 及初始条件 y(0) = a，反复对原方程求导可逐步得出 y'(0), y”(0), y”'(0) 等，从而构建级数解。这是绕开解析求解的一个灵巧应用。
A typical task: ‘Find the first four terms of the Maclaurin series for y, given that y’ = x + y and y(0) = 1.’ Start with y(0)=1, then y'(0)=0+1=1. Differentiate: y” = 1 + y’ ⇒ y”(0) = 1+1=2. Next, y”’ = y” ⇒ y”'(0)=2. Hence y ≈ 1 + 1x + 2x²/2! + 2x³/3! = 1 + x + x² + x³/3. Always state each step clearly.
典型题：“已知 y’ = x + y 且 y(0) = 1，求 y 的麦克劳林级数的前四项。” 从 y(0)=1 出发，y'(0)=0+1=1。求导：y” = 1 + y’ ⇒ y”(0)=1+1=2。再求导：y”’ = y” ⇒ y”'(0)=2。因此 y ≈ 1 + 1x + 2x²/2! + 2x³/3! = 1 + x + x² + x³/3。每一步都要清晰呈现。
9. Common Pitfalls and How to Avoid Them | 常见错误与避坑指南
One frequent mistake is forgetting to divide by the factorial when writing coefficients. The term in x³ is f”'(0)/6, not just f”'(0). Double-check each term before moving on. Another is mis-calculating the derivative at zero, especially with products or chain rules. Always evaluate carefully and watch signs.
一个常见错误是写出系数时忘记除以阶乘。x³ 项是 f”'(0)/6，而不是直接写 f”'(0)。每写完一项都应复核。另一个问题是求 x=0 处导数值时出错，尤其是涉及乘法法则或链式法则时。务必仔细计算，注意符号。
Also, many students ignore the validity statement or write it incorrectly. Even if the question only asks for the series up to x³, a validity note (e.g. for all real x, or for |x|<1) is often required. Read the command words: 'state the range of validity' means you must write this.
此外，不少考生忽略了有效性的表述，或写错。即使题目只要求写到 x³ 项，通常也要求附上有效性（如对所有实数 x，或 |x|<1）。务必看清指令词：“写出有效范围”即须明确给出。
Finally, in substitution problems, students sometimes fail to expand brackets properly when powers of x are combined. For example, when substituting u = 2x into (1+u)ⁿ, the term (2x)² is 4x², not 2x². A classic slip that costs marks.
最后，在代换题中，学生有时展开括号时出错，尤其在 x 的幂组合时。例如将 u=2x 代入 (1+u)ⁿ，(2x)² 是 4x² 而非 2x²。这是个会丢分的经典笔误。
10. Exam-Style Worked Example | 考试题型实例精讲
Question: Find the Maclaurin series for f(x) = ln(1+ sin x) up to and including the term in x³. State any restrictions on x.
题目：求 f(x) = ln(1+ sin x) 的麦克劳林级数，含 x³ 项为止，并指出 x 的限制条件。
Solution: Use known series. sin x = x − x³/6 + … (all x). Let u = sin x, then ln(1+u) = u − u²/2 + u³/3 − … for −1< u ≤1. Substitute u = x − x³/6 + ...:
解：使用已知级数。sin x = x − x³/6 + …（对所有 x）。令 u = sin x，则 ln(1+u) = u − u²/2 + u³/3 − …（−1< u ≤1）。代入 u = x − x³/6 + ...：
u = x − x³/6
u² = (x − x³/6)² = x² − x⁴/3 + … → keep only x² (x⁴ discarded)
u³ = (x − x³/6)³ = x³ − 3x⁵/6 + … → keep x³
Now assemble: ln(1+sin x) = (x − x³/6) − ½(x²) + ⅓(x³) + … = x − x²/2 + ( −1/6 + 1/3 ) x³ = x − ½x² + ⅙x³ + …
Validity: For the series to be valid, we need −1 < sin x ≤ 1. sin x ≤ 1 always holds; sin x > −1 is violated only at isolated points (where x = −π/2 + 2kπ), so for small x near 0 the series is valid for |x| small enough, typically within the radius of convergence of the full series. A safe statement: ‘valid for all real x’ is not strictly true because ln(1+sinx) is undefined when sin x = −1. But for the Maclaurin expansion, we take the function as defined near 0, so the series expansion is valid in some interval around 0. OCR typically accepts ‘valid for x in a sufficiently small interval containing 0’ or you can leave as ‘for |sin x| < 1, i.e. x ≠ −π/2 + 2πk' . In practice, for the polynomial approximation, you may state the obvious that the series is valid for |x| small enough such that |sin x| < 1, which is all x except those where sin x = −1. A compact exam answer: 'approximately true for x near 0'.
有效性讨论：原级数要求 −1 < sin x ≤ 1。sin x ≤ 1 恒成立，仅当 sin x = −1 时级数发散。因此该级数在包含 0 的某个区间内收敛，可写“对于充分小的 x 有效”。在考试中通常可注明“x 接近 0 时成立”或“在 |x| 很小时有效”。
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