Edexcel Maths: Taylor Series Key Points Explained | 爱德思数学：泰勒级数考点精讲

📚 Edexcel Maths: Taylor Series Key Points Explained | 爱德思数学：泰勒级数考点精讲

Taylor series provide a powerful way to represent a wide class of functions as infinite sums of polynomial terms. In Edexcel A-Level Further Mathematics, you are expected to derive standard Maclaurin expansions, determine intervals of convergence, use series for approximations, and solve related differential equations.

泰勒级数是一种将众多函数表示为无穷多项式之和的强大工具。在爱德思 A-Level 进阶数学中，你需要掌握推导标准麦克劳林展开式、确定收敛区间、利用级数进行近似计算以及求解相关微分方程。

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1. Definition of Taylor Series | 泰勒级数的定义

For a function f(x) that is infinitely differentiable at x = a, the Taylor series expansion about a is f(a) + f'(a)(x – a) + f”(a)(x – a)²/2! + f”'(a)(x – a)³/3! + … .

如果一个函数 f(x) 在 x = a 处无穷可导，它在 a 处的泰勒级数展开为 f(a) + f'(a)(x – a) + f”(a)(x – a)²/2! + f”'(a)(x – a)³/3! + … 。

The series can be written compactly using sigma notation: ∑ₙ₌₀^∞ f⁽ⁿ⁾(a) (x – a)ⁿ / n!.

该级数可用求和符号简洁表示为：∑ₙ₌₀^∞ f⁽ⁿ⁾(a) (x – a)ⁿ / n!。

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2. Maclaurin Series as a Special Case | 麦克劳林级数作为特例

When a = 0, the Taylor series reduces to a Maclaurin series: f(x) = f(0) + f'(0)x + f”(0)x²/2! + f”'(0)x³/3! + … . This is the most common type you will work with in exams.

当 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! + … 。这是你在考试中最常处理的类型。

Edexcel questions almost always ask you to find the Maclaurin expansion up to a specific term, such as x⁴ or x⁵, so you must be comfortable computing derivatives at 0.

爱德思试题几乎总是要求你求出麦克劳林展开到某一特定项，如 x⁴ 或 x⁵，因此你必须能够熟练计算在 0 处的导数。

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3. Standard Maclaurin Expansions to Memorise | 必记的标准麦克劳林展开

The exam expects instant recall of these series (with their validity intervals):

考试要求能立即写出这些级数（连同其有效区间）：

• eˣ = 1 + x + x²/2! + x³/3! + … , for all x (all real numbers)

  eˣ = 1 + x + x²/2! + x³/3! + … ，对所有实数 x 成立

• sin x = x – x³/3! + x⁵/5! – … , for all x

  sin x = x – x³/3! + x⁵/5! – … ，对所有 x 成立

• cos x = 1 – x²/2! + x⁴/4! – … , for all x

  cos x = 1 – x²/2! + x⁴/4! – … ，对所有 x 成立

• ln(1 + x) = x – x²/2 + x³/3 – x⁴/4 + … , for -1 < x ≤ 1

  ln(1 + x) = x – x²/2 + x³/3 – x⁴/4 + … ，有效区间 -1 < x ≤ 1

• (1 + x)ⁿ = 1 + nx + n(n-1)x²/2! + … , for |x| < 1 (the binomial series)

  (1 + x)ⁿ = 1 + nx + n(n-1)x²/2! + … ，有效区间 |x| < 1（二项式级数）

You may also need to derive series for sec x, tan x, or arcsin x using composition or division, but these core five form the backbone of the topic.

你可能还需要用复合或除法推导 sec x、tan x 或 arcsin x 的级数，但这五个核心级数是本专题的基础。

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4. Validity Range and Convergence | 有效范围与收敛性

Every Maclaurin series has a range of x-values for which the infinite sum equals the original function. For eˣ, sin x, and cos x the series converge for all real x.

每个麦克劳林级数都有一个使无穷和等于原函数的 x 取值范围。对于 eˣ、sin x 和 cos x，其级数对所有实数 x 收敛。

For ln(1 + x) the series is valid for -1 < x ≤ 1; for the binomial series with exponent n, it converges for |x| < 1 unless n is a positive integer, in which case the series terminates.

对于 ln(1 + x)，级数在 -1 < x ≤ 1 有效；对于指数为 n 的二项式级数，收敛范围为 |x| < 1，除非 n 为正整数（此时级数有限）。

You must state the validity range when asked for an expansion; otherwise, marks are lost.

当题目要求展开时，你必须写明有效范围，否则将失分。

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5. Term-by-term Differentiation and Integration | 逐项求导与积分

A series can be differentiated or integrated term by term within its interval of convergence. This is useful for obtaining new series from known ones.

级数在其收敛区间内可以逐项求导或积分。这有助于从已知级数获得新的级数。

For example, differentiating the series for sin x term by term gives the series for cos x: (x – x³/3! + … )’ = 1 – x²/2! + … .

例如，对 sin x 的级数逐项求导即可得到 cos x 的级数：(x – x³/3! + … )’ = 1 – x²/2! + … 。

Similarly, integrating the geometric series 1/(1 + x) = 1 – x + x² – x³ + … (|x| < 1) term by term produces the series for ln(1 + x).

类似地，将几何级数 1/(1 + x) = 1 – x + x² – x³ + … (|x| < 1) 逐项积分，即可得到 ln(1 + x) 的级数。

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6. Using Substitutions to Derive New Series | 使用代换推导新级数

A fast method to expand a function like e³ˣ or sin(2x) is to replace x in the standard series. For e³ˣ, simply substitute 3x for x in the eˣ series: 1 + (3x) + (3x)²/2! + … = 1 + 3x + 9x²/2 + … .

展开诸如 e³ˣ 或 sin(2x) 这类函数的快捷方法是将标准级数中的 x 替换掉。对于 e³ˣ，只需在 eˣ 级数中用 3x 替换 x：1 + (3x) + (3x)²/2! + … = 1 + 3x + 9x²/2 + … 。

The new validity range follows from the substitution. Since eˣ is valid for all x, e³ˣ is also valid for all x. For ln(1 + 2x), the condition -1 < 2x ≤ 1 gives -½ < x ≤ ½.

新的有效范围由代换决定。因为 eˣ 对所有 x 有效，故 e³ˣ 也对所有 x 有效。对于 ln(1 + 2x)，条件 -1 < 2x ≤ 1 得到 -½ < x ≤ ½。

You can also combine expansions: e.g., multiply the series for eˣ and cos x up to a certain order to find the expansion of eˣcos x.

你也可以组合展开式，例如将 eˣ 和 cos x 的级数乘到某一阶，以求出 eˣcos x 的展开式。

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7. Approximating Functions Using Taylor Polynomials | 利用泰勒多项式近似函数

Truncating the series after a few terms gives a Taylor polynomial that approximates the function near x = a. The more terms you include, the better the approximation.

在几项之后截断级数，可得到一个泰勒多项式，用于近似函数在 x = a 附近的值。包含的项越多，近似越精确。

A typical exam problem: use the expansion of √(1 + 2x) up to x² to estimate √1.02, then find the percentage error. You will need to set 2x = 0.02, so x = 0.01.

典型的考题：利用 √(1 + 2x) 到 x² 项的展开式估计 √1.02，然后计算百分比误差。此时需令 2x = 0.02，故 x = 0.01。

Remember that the approximation is only valid if x lies within the interval of convergence.

请记住，只有当 x 处于收敛区间内时，这种近似才有效。

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8. Error Estimation (Lagrange Remainder) | 拉格朗日余项与误差估计

The Lagrange form of the remainder term Rₙ(x) = f⁽ⁿ⁺¹⁾(c) xⁿ⁺¹/(n+1)! for some c between 0 and x gives the maximum possible error when truncating a Maclaurin series to n+1 terms.

拉格朗日余项形式 Rₙ(x) = f⁽ⁿ⁺¹⁾(c) xⁿ⁺¹/(n+1)!（其中 c 介于 0 和 x 之间）给出了将麦克劳林级数截断到 n+1 项时的最大可能误差。

For example, to find how many terms of eˣ series are needed to approximate e to within 0.001, you can bound the derivative term |f⁽ⁿ⁺¹⁾(c)| ≤ e (since eˣ ≤ e for x=1) and solve the inequality.

例如，要确定需要 eˣ 级数的多少项才能使 e 的近似值误差小于 0.001，可以限制导数项 |f⁽ⁿ⁺¹⁾(c)| ≤ e（因为 x=1 时 eˣ ≤ e），然后解不等式。

Although not always tested, Edexcel FP2 can include a question on bounding the error, so be prepared.

虽然不常考，但爱德思 FP2 可能包含估计误差的题目，要做好准备。

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9. Solving Differential Equations with Series | 用级数求解微分方程

When a differential equation cannot be solved by standard methods, you may be asked to find a series solution. Assume y = a₀ + a₁x + a₂x² + a₃x³ + … , substitute into the equation, and equate coefficients.

当微分方程不能用标准方法求解时，可能会要求你求级数解。设 y = a₀ + a₁x + a₂x² + a₃x³ + … ，代入方程并匹配系数。

This process yields recurrence relations that allow you to determine coefficients step by step, up to the required term.

这一过程会得到递推关系，使你能够逐步求出满足要求项的系数。

A classic Edexcel question: solve y” + x y’ + y = 0 by a Maclaurin series up to x⁴, given y(0) = 1, y'(0) = 0.

一道经典的爱德思考题为：已知 y(0) = 1，y'(0) = 0，用麦克劳林级数求 y” + x y’ + y = 0 到 x⁴ 为止的解。

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10. Limits and Asymptotic Behaviour Using Series | 利用级数求极限与渐近行为

Series are a handy tool for evaluating limits that produce indeterminate forms. Expanding numerator and denominator as Maclaurin series often cancels troublesome terms and reveals the limit.

级数是计算不定型极限的便捷工具。将分子和分母展开为麦克劳林级数，常常能约去令人头疼的项并揭示极限值。

For example, limₓ→₀ (eˣ – 1 – x)/x² can be found by substituting eˣ = 1 + x + x²/2 + … , giving (x²/2 + …)/x² → ½ as x → 0.

例如，limₓ→₀ (eˣ – 1 – x)/x² 可通过代入 eˣ = 1 + x + x²/2 + … 求得，化简得 (x²/2 + …)/x² → ½（x → 0）。

This technique regularly appears in Further Pure exams and can save you from messy L’Hopital’s rule applications.

该技巧在进阶纯数考试中时常出现，能省去繁琐的洛必达法则应用。

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11. Common Pitfalls and Exam Strategies | 常见陷阱与应试策略

Many candidates lose marks by forgetting the factorial denominators or misplacing signs in the expansions of sin and cos. Always double-check each term.

许多考生因忘记阶乘分母，或在 sin 和 cos 展开中错放符号而丢分。务必逐项复查。

When the question says ‘up to and including x⁴’, include the x⁴ term fully; if it says ‘as far as x³’, stop after x³. Sloppy reading costs marks.

当题目要求“直到并包括 x⁴”时，需完整包含 x⁴ 项；若要求“到 x³ 为止”，则在 x³ 后停止。审题粗心会导致失分。

Always state the validity range at the end of the expansion, even if the question only implies it. Use the notation |x| < 1 or -1 < x ≤ 1 precisely.

即使题目仅是隐晦提及，也务必在展开式末尾写明有效范围，并精确使用 |x| < 1 或 -1 < x ≤ 1 等符号。

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12. Summary Table of Key Expansions | 关键展开式总结表

Function	Maclaurin series (first 4 terms)	Validity
eˣ	1 + x + x²/2! + x³/3! + …	all real x
sin x	x – x³/3! + x⁵/5! – x⁷/7! + …	all real x
cos x	1 – x²/2! + x⁴/4! – x⁶/6! + …	all real x
ln(1 + x)	x – x²/2 + x³/3 – x⁴/4 + …	-1 < x ≤ 1
(1 + x)ⁿ	1 + nx + n(n-1)x²/2! + n(n-1)(n-2)x³/3! + …	|x| < 1 (if n not positive integer)

Memorising this table and practicing the associated skills will give you confidence in tackling any Taylor series question on the Edexcel paper.

记熟此表并勤加练习相关技能，将使你有信心应对爱德思考卷中任何泰勒级数考题。

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