📚 Taylor Series for OCR A-Level Mathematics | A-Level OCR 数学：泰勒级数考点精讲
Taylor series is a cornerstone of A-Level Further Mathematics under OCR, allowing you to represent infinitely differentiable functions as power series centred at a point. In the exam, you will be expected to derive Maclaurin series, manipulate standard expansions, use them for approximations, evaluate limits, and understand the error involved. A solid grasp of differentiation, factorial notation, and convergence is essential to excel in this topic.
泰勒级数是 OCR 进阶数学的重要基石，它把无限可微的函数表示为以某一点为中心的幂级数。考试中，你需要推导麦克劳林级数、处理标准展开式、用级数做近似计算、求极限，并理解相关的误差。扎实掌握微分、阶乘符号和收敛性，是在这部分取得高分的关键。
1. What is a Taylor Series? | 什么是泰勒级数？
Given a function f(x) that is infinitely differentiable at x = a, its Taylor series expansion about a is f(a) + f'(a)(x – a) + f”(a)(x – a)2/2! + f”'(a)(x – a)3/3! + … . In compact form, the series is written as ∑n=0∞ f(n)(a) (x – a)n / n!, where f(n)(a) denotes the n-th derivative evaluated at a.
若函数 f(x) 在 x = a 处无限可微，则它在该点附近的泰勒级数展开为 f(a) + f'(a)(x – a) + f”(a)(x – a)2/2! + f”'(a)(x – a)3/3! + …。紧凑形式写作 ∑n=0∞ f(n)(a) (x – a)n / n!，其中 f(n)(a) 表示在 a 处的 n 阶导数值。
The finite sum up to the n-th term is called the Taylor polynomial of degree n, which provides a local polynomial approximation to f(x) near x = a. As more terms are added, the approximation improves provided the series converges.
有限项求和到第 n 项称为 n 次泰勒多项式，它为 x = a 附近的 f(x) 提供了一个局部多项式逼近。增加更多项，只要级数收敛，逼近效果就会更好。
2. Maclaurin Series: The Special Case a = 0 | 麦克劳林级数：a = 0 的特殊情形
A Maclaurin series is simply a Taylor series centred at a = 0. For OCR exams, most questions focus on Maclaurin expansions because they simplify calculations and match the standard series for common functions. The general Maclaurin series is f(0) + f'(0)x + f”(0)x2/2! + f”'(0)x3/3! + … = ∑n=0∞ f(n)(0) xn / n!.
麦克劳林级数就是中心在 a = 0 的泰勒级数。OCR 考试大部分题目集中在麦克劳林展开，因为计算更简单，而且与常见函数的标准级数一致。其一般形式为 f(0) + f'(0)x + f”(0)x2/2! + f”'(0)x3/3! + … = ∑n=0∞ f(n)(0) xn / n!。
To find a Maclaurin series, you repeatedly differentiate the function, evaluate each derivative at x = 0, substitute into the formula, and identify the pattern for the general term. Remember to check that the function is well-behaved at 0.
求麦克劳林级数的方法是：反复对函数求导，在 x = 0 处计算各阶导数值，代入公式，并找出通项的模式。注意要确保函数在 0 处性质良好。
3. Step-by-Step: Building a Maclaurin Series | 逐步建立麦克劳林级数
Let us take f(x) = e2x as an example. First, list derivatives: f(x) = e2x, f'(x) = 2e2x, f”(x) = 4e2x, f”'(x) = 8e2x, … f(n)(x) = 2n e2x. Then evaluate at 0: f(n)(0) = 2n.
以 f(x) = e2x 为例。先列出各阶导数：f(x) = e2x, f'(x) = 2e2x, f”(x) = 4e2x, f”'(x) = 8e2x, … f(n)(x) = 2n e2x。再在 0 处求值：f(n)(0) = 2n。
Now plug into the series: e2x = 1 + 2x + (4x2)/2! + (8x3)/3! + … = 1 + 2x + 2x2 + (4/3)x3 + … . The general term is (2n/n!) xn. Always simplify coefficients where possible.
代入级数公式：e2x = 1 + 2x + (4x2)/2! + (8x3)/3! + … = 1 + 2x + 2x2 + (4/3)x3 + …。通项为 (2n/n!) xn。尽可能化简系数。
4. Standard Maclaurin Series You Must Memorise | 必须记住的标准麦克劳林级数
OCR expects you to know these expansions and their intervals of validity off by heart. The table below summarises the five most important series.
OCR 要求你熟记以下展开式及其有效区间。下表总结了最重要的五个级数。
| f(x) | Maclaurin series | Valid for |
| ex | 1 + x + x2/2! + x3/3! + … = ∑n=0∞ xn/n! | all real x |
| sin x | x – x3/3! + x5/5! – … = ∑n=0∞ (-1)n x2n+1/(2n+1)! | all real x |
| cos x | 1 – x2/2! + x4/4! – … = ∑n=0∞ (-1)n x2n/(2n)! | all real x |
| ln(1 + x) | x – x2/2 + x3/3 – … = ∑n=1∞ (-1)n-1 xn/n | -1 < x ≤ 1 |
| (1 + x)k | 1 + kx + k(k-1)x2/2! + k(k-1)(k-2)x3/3! + … | |x| < 1 (unless k ∈ ℕ) |
Note that for (1 + x)k, the expansion is finite when k is a positive integer, giving the binomial theorem. For non-integer k, the series is infinite and converges only for |x| < 1.
注意，当 k 为正整数时，(1 + x)k 的展开式为有限项（二项式定理）；若 k 为非整数，则级数无穷且仅当 |x| < 1 时收敛。
5. Manipulating Series: Substitution and Composition | 级数变形：代换与复合
You can generate new expansions from the standard ones by substituting into x. For example, replacing x by -x in ex gives e-x = 1 – x + x2/2! – x3/3! + … . To expand sin(3x), replace x by 3x in the sin x series: sin(3x) = 3x – (3x)3/3! + (3x)5/5! – … .
你可以通过对标准级数中的 x 进行代换来获得新展开式。例如，将 ex 中的 x 换为 -x 得到 e-x = 1 – x + x2/2! – x3/3! + …。要展开 sin(3x)，把 sin x 级数中的 x 替换为 3x：sin(3x) = 3x – (3x)3/3! + (3x)5/5! – …。
Composition is another powerful tool. To find the series for ex², substitute x² into the ex series: ex² = 1 + x² + x⁴/2! + x⁶/3! + … . This is valid for all x because ex converges everywhere. When combining series, always ensure the final expression is correctly simplified and that the convergence interval adapts accordingly.
复合是另一个有力工具。求 ex² 的级数，将 x² 代入 ex 级数：ex² = 1 + x² + x⁴/2! + x⁶/3! + …。这对所有实数 x 有效，因为 ex 处处收敛。组合级数时，务必确保最终表达式正确化简，并注意收敛区间相应调整。
Multiplication by a polynomial or integration/differentiation of a series can also be required. For instance, to obtain arctan x, integrate the geometric series 1/(1 + x²) = 1 – x² + x⁴ – … term-by-term to get x – x³/3 + x⁵/5 – …, valid for |x| < 1.
考试还可能涉及级数乘多项式、对级数逐项积分或求导。例如，对几何级数 1/(1 + x²) = 1 – x² + x⁴ – … 逐项积分得到 arctan x ≈ x – x³/3 + x⁵/5 – …，有效区间为 |x| < 1。
6. Convergence and Interval of Validity | 收敛性与有效区间
A Maclaurin series is only equal to the function where it converges. You must be able to state the interval of convergence, typically found using the ratio test or from known standard series. For ex, sin x, cos x, the series converges for all real x. For ln(1 + x), the interval is -1 < x ≤ 1, and for (1 + x)k with non-integer k, it is |x| < 1.
麦克劳林级数仅在收敛域内等于原函数。你需要能够说明收敛区间，常用比值判别法或基于已知标准级数给出。eˣ、sin x、cos x 的级数对所有实数收敛；ln(1 + x) 的收敛区间为 -1 < x ≤ 1；非整数指数 (1 + x)k 的收敛域为 |x| < 1。
In OCR questions, you may be asked to find the first few non-zero terms and state the range of values of x for which the expansion is valid. Always check endpoints unless the question explicitly asks for the open interval.
OCR 题目可能要求给出最初几个非零项，并说明展开式有效的 x 取值范围。除非题目明确要求开区间，否则要检查端点情况。
7. Lagrange Remainder and Error Estimation | 拉格朗日余项与误差估计
When you truncate a Taylor series after n terms, the error Rn(x) can be bounded using the Lagrange form: Rn(x) = f(n+1)(c) (x – a)n+1 / (n+1)! for some c between a and x. For Maclaurin series, a = 0. This remainder estimate tells you how many terms are needed to achieve a desired accuracy in approximation.
将泰勒级数截断为 n 项后，误差 Rn(x) 可用拉格朗日余项估计：Rn(x) = f(n+1)(c) (x – a)n+1 / (n+1)!，其中 c 介于 a 与 x 之间。对麦克劳林级数，a = 0。这一余项估计告诉你需要多少项才能达到预定的近似精度。
A typical OCR question might ask: “Using the Maclaurin series for cos x, estimate cos 0.2 to 4 decimal places and justify the number of terms used.” You would bound the next term containing the maximum possible derivative value and ensure it is less than 5×10⁻⁵.
典型的 OCR 考题可能这样问：“利用 cos x 的麦克劳林级数估计 cos 0.2 到 4 位小数，并论证所用项数。”你需要用导数最大值去界定下一项，并确保该项小于 5×10⁻⁵。
8. Applications: Limits and Approximations | 应用：极限与近似计算
Series expansions simplify evaluation of limits that produce indeterminate forms like 0/0. For example, limx→0 (sin x – x)/x³ can be evaluated by substituting sin x = x – x³/3! + … , giving limit -1/6. This method is often quicker than l’Hopital’s rule.
级数展开能简化产生 0/0 等不定式极限的计算。例如，limx→0 (sin x – x)/x³，代入 sin x = x – x³/3! + … 可直接得到极限 -1/6。此法常比洛必达法则更快捷。
For approximations, using the first two or three terms often suffices for small x. Estimate √(1.02) by writing (1 + 0.02)½ ≈ 1 + ½(0.02) – ⅛(0.02)² + … = 1 + 0.01 – 0.00005 = 1.00995, which is correct to five decimal places.
在近似计算中，对较小的 x 通常取前两三项即可。要估计 √(1.02)，可写作 (1 + 0.02)½ ≈ 1 + ½(0.02) – ⅛(0.02)² + … = 1 + 0.01 – 0.00005 = 1.00995，精确到五位小数。
9. Series for Integrals: Termwise Integration | 级数积分：逐项积分法
When a function cannot be integrated in closed form, its series enables numerical integration. For example, ∫₀0.5 e-x² dx can be approximated by expanding e-x² = 1 – x² + x⁴/2! – x⁶/3! + … and integrating termwise: [x – x³/3 + x⁵/(5·2!) – x⁷/(7·3!) + …] evaluated from 0 to 0.5. The result approximates 0.461.
当函数无法用初等函数积分时，它的级数可实现数值积分。例如，∫₀0.5 e-x² dx 可先展开 e-x² = 1 – x² + x⁴/2! – x⁶/3! + …，再逐项
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