Taylor Series: CIE A-Level Maths Key Points | 泰勒级数 考点精讲

📚 Taylor Series: CIE A-Level Maths Key Points | 泰勒级数 考点精讲

The Taylor series provides a powerful way to represent a function as an infinite sum of terms calculated from its derivatives at a single point. It is a cornerstone of A-Level Pure Mathematics, enabling polynomial approximations and elegant evaluations of limits and integrals. In CIE exams, you are expected to derive and apply Maclaurin series for standard functions, manipulate expansions, and understand their ranges of validity.

泰勒级数利用函数在某一点的各阶导数构造一个无穷级数，从而实现用多项式逼近复杂函数。它是 A-Level 纯数的重要工具，能帮助解决极限、积分近似等问题。CIE 考试要求掌握标准函数的麦克劳林级数推导、展开式的变形以及收敛区间的判断。

1. Introduction to Taylor Series | 泰勒级数简介

The Taylor series of a function f(x) about a point x = a expresses the function as an infinite sum involving its derivatives at a. The general form is:

函数 f(x) 在点 x = a 的泰勒级数用该点处的各阶导数构造一个无穷级数，其一般表达式为：

f(x) = Σ (from n=0 to ∞) [f⁽ⁿ⁾(a)/n!] (x − a)ⁿ

Here f⁽ⁿ⁾(a) denotes the n-th derivative evaluated at a, and n! is the factorial. The sum turns the function into a polynomial of infinite degree, which often converges to f(x) for x near a. In CIE, you will mostly work with a special case where a = 0.

其中 f⁽ⁿ⁾(a) 表示 f 在 a 处的 n 阶导数，n! 为阶乘。这个无穷多项式在 a 附近往往收敛到原函数。CIE 考试中主要考察 a = 0 的特例。

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2. Maclaurin Series: The Special Case a = 0 | 麦克劳林级数：a = 0 的特例

When the expansion is centred at the origin (a = 0), the Taylor series simplifies to the Maclaurin series:

当展开中心取为原点 (a = 0) 时，泰勒级数简化为 麦克劳林级数：

f(x) = f(0) + f'(0)x + f”(0)x²/2! + f”'(0)x³/3! + …

This is the form required by the CIE 9709 syllabus. You must be able to compute several derivatives, evaluate them at 0, and write down the series up to a required term, typically up to x³ or x⁴.

这是 CIE 9709 大纲要求的重点形式。考生需能够求出若干阶导数、计算它们在 0 处的值，并按题目要求写出级数，通常写到 x³ 或 x⁴ 项。

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

Let f(x) = eˣ. All derivatives are eˣ, so at x = 0 every derivative equals 1. Substituting into the Maclaurin formula gives:

设 f(x) = eˣ。所有导数都是 eˣ，因此在 x = 0 处各阶导数值均为 1。代入麦克劳林公式得到：

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

This series converges for all real x. It is one of the most frequently tested expansions; you should memorise it and be able to use it for substitutions such as e²ˣ or e⁻ˣ.

该级数对所有实数 x 都收敛。这是考试中最高频的展开式之一，必须熟记并能灵活用于代换，如 e²ˣ 或 e⁻ˣ。

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4. Series for Sine and Cosine | 正弦与余弦的级数展开

The trigonometric functions sin x and cos x have derivatives that cycle every four steps. At x = 0: sin(0) = 0, sin'(0) = 1, sin”(0) = 0, sin”'(0) = −1, etc. The resulting series are:

正弦和余弦函数的导数呈现四步循环。在 x = 0 处：sin(0) = 0，sin'(0) = 1，sin”(0) = 0，sin”'(0) = −1，等等。由此得到：

sin x = x − x³/3! + x⁵/5! − x⁷/7! + …

cos x = 1 − x²/2! + x⁴/4! − x⁶/6! + …

Both series are valid for all real x. Notice the alternating signs and only odd powers for sin, even powers for cos. These patterns are easy to recall and often appear in approximation questions.

这两个级数对所有实数均成立。注意正弦展开只含奇次幂，余弦展开只含偶次幂，且正负号交替。这些规律便于记忆，常出现在近似计算题中。

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5. Expansion of ln(1 + x) and Its Validity | ln(1 + x) 的展开及有效区间

For f(x) = ln(1 + x), we have f(0) = 0, f‘(x) = (1+x)⁻¹, f”(x) = −(1+x)⁻², f”'(x) = 2(1+x)⁻³. Evaluating at 0 yields a pattern that gives:

对于 f(x) = ln(1 + x)，有 f(0) = 0，f‘(x) = (1+x)⁻¹，f”(x) = −(1+x)⁻²，f”'(x) = 2(1+x)⁻³。代入 0 得到规律：

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

The expansion is valid for −1 < x ≤ 1. At x = 1, the series becomes the alternating harmonic series, which converges conditionally; at x = −1, ln(0) is undefined. The CIE mark scheme often requires you to state the interval of validity explicitly.

该展开式有效区间为 −1 < x ≤ 1。当 x = 1 时级数化为交错调和级数，条件收敛；而 x = −1 时函数无定义。CIE 评分标准通常要求明确写出收敛区间。

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6. Binomial Expansion as a Maclaurin Series | 二项式展开作为麦克劳林级数

The binomial theorem for (1 + x)ⁿ, where n is any real number, can be derived as a Maclaurin series:

二项式展开 (1 + x)ⁿ（n 为任意实数）可视为麦克劳林级数：

(1 + x)ⁿ = 1 + n x + [n(n−1)/2!] x² + [n(n−1)(n−2)/3!] x³ + …

When n is a positive integer, the series terminates and is exact for all x. For non-integer n, the series is infinite and converges only for |x| < 1. This is a direct consequence of the Maclaurin formula and is extremely useful for expanding expressions like √(1+x) or (1+x)⁻¹.

当 n 为正整数时级数有限，对所有 x 成立；当 n 不是正整数时，级数无限且仅在 |x| < 1 收敛。这一展开式在化简如 √(1+x) 或 (1+x)⁻¹ 时非常实用。

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7. Composing Series: Substitution and Multiplication | 级数的复合：代换与乘法

You rarely need to differentiate from first principles for every function. Instead, you can substitute into known series. For instance, to expand e³ˣ, replace x by 3x in the series for eˣ:

实际考试中很少每次都从导数开始推导，可使用已知级数进行代换。例如展开 e³ˣ，只需将 eˣ 级数中的 x 替换为 3x：

e³ˣ = 1 + 3x + (9x²)/2! + (27x³)/3! + …

Multiplication of series is also examined. To find the Maclaurin series for eˣ sin x up to x³, multiply the two expansions and collect like terms: (1 + x + x²/2 + x³/6)(x − x³/6) = x + x² + (1/3)x³ + … . Always truncate to the required order.

级数乘法也常考。如求 eˣ sin x 的麦克劳林级数到 x³，将两展开式相乘并合并同类项：(1 + x + x²/2 + x³/6)(x − x³/6) = x + x² + (1/3)x³ + …。注意只需取到所需最高次幂。

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8. Approximating Functions and Integrals | 函数值与积分的近似计算

Truncating a Maclaurin series after a few terms gives a polynomial that approximates the function near x = 0. For example, to estimate e⁰·¹:

截断麦克劳林级数得到一个多项式，可在 x = 0 附近近似原函数。例如估计 e⁰·¹：

e⁰·¹ ≈ 1 + 0.1 + (0.1)²/2 + (0.1)³/6 = 1.1051666…

Similarly, integrals such as ∫₀⁰·² e⁻ˣ² dx can be approximated by first expanding e⁻ˣ² as 1 − x² + x⁴/2 − …, integrating term by term, and then evaluating. This technique appears regularly in CIE P3 and Further Maths papers.

类似地，对于积分 ∫₀⁰·² e⁻ˣ² dx，可先将 e⁻ˣ² 展开为 1 − x² + x⁴/2 − …，然后逐项积分并求值。这种方法在 CIE P3 及进阶数学卷中很常见。

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9. Error Estimation: The Lagrange Remainder | 误差估计：拉格朗日余项

When you use only the first few terms of a Taylor/Maclaurin series, the remainder Rₙ gives the error. The Lagrange form states that for some c between a and x:

当你只取泰勒/麦克劳林级数的前几项时，剩余项 Rₙ 代表了误差。拉格朗日余项形式为，在 a 与 x 之间存在某个 c 使得：

Rₙ = f⁽ⁿ⁺¹⁾(c) / (n+1)! × (x − a)ⁿ⁺¹

For alternating series that satisfy the conditions of the alternating series test, the absolute error is less than the magnitude of the first omitted term. This simple bound is often sufficient in CIE problems to justify the accuracy of an approximation.

对于满足交错级数判别的级数，绝对误差不超过首个被忽略项的绝对值。CIE 试题常利用这个简单界限来论证近似值的精度。

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10. Common Exam Pitfalls and Tips | 常见易错点与答题技巧

Students frequently lose marks by forgetting to divide by the factorial, especially when computing beyond the second derivative. Always write down the general term or at least check that your coefficients match 1/2! or 1/3! rather than just 1/2 or 1/3. Also, when a question asks for the ‘series expansion’, you must state the range of validity if it is not the whole real line.

考生常因忘记除以阶乘而丢分，尤其是在求三阶以上导数时。务必确认系数含有 1/2!、1/3! 等，而非单纯的 1/2 或 1/3。此外，题目要求写出“级数展开”时，若收敛区间并非全体实数，必须同时标明有效范围。

When expanding a composite function like ln(1 + 2x), do not simply substitute into the ln(1 + x) series and forget to adjust the validity interval: the expansion for ln(1 + 2x) is valid for −1 < 2x ≤ 1, i.e. −½ < x ≤ ½. Similarly, watch for products and series: always multiply out and collect terms up to the power required.

对复合函数如 ln(1 + 2x) 展开时，不要只机械代入而忘了调整有效区间：ln(1 + 2x) 的展开成立当 −1 < 2x ≤ 1，即 −½ < x ≤ ½。乘积级数也一样，相乘后需整理到题目要求的最高次幂。

Finally, practice differentiating accurately: one sign error in the third derivative can ruin the entire expansion. Use a systematic approach and double-check your derivatives before substituting x = 0.

最后，务必练熟求导：三阶导数的一个符号错误就可能毁掉整个展开式。按步骤计算并在代入 x = 0 前仔细核对导数结果。

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