Maclaurin Series: Key Exam Points | 麦克劳林展开考点精讲

📚 Maclaurin Series: Key Exam Points | 麦克劳林展开考点精讲

Maclaurin series is a core topic in IB AQA Mathematics, linking differentiation, sequences, and function approximation. Mastering this topic requires understanding the derivation, memorising standard expansions, and applying techniques such as substitution, differentiation, and multiplication to find new series. This guide breaks down every essential exam point, from basic definitions to Lagrange remainder error bounds, with clear explanations in both English and Chinese.

麦克劳林级数是 IB AQA 数学的核心内容，它把求导、数列和函数近似联系在一起。掌握这一主题需要理解推导过程、熟记标准展开式，并灵活运用代入、求导和乘法等技巧求出新的级数。本文从基本定义到拉格朗日余项误差界限，用中英双语逐一剖析每一个重要的考点。

1. What is a Maclaurin Series? | 什么是麦克劳林级数？

A Maclaurin series is a Taylor series expansion of a function about x = 0. It represents an infinitely differentiable function f(x) as an infinite sum of terms calculated from the values of its derivatives at zero.

麦克劳林级数是函数在 x = 0 处的泰勒级数展开。它把一个无限可导的函数 f(x) 表示为一个无穷求和，每一项都由函数在 0 处的各阶导数值计算得到。

The general formula is:

通项公式为：

f(x) = f(0) + f′(0)x + f″(0)x²/2! + f‴(0)x³/3! + … + f⁽ⁿ⁾(0)xⁿ/n! + …

In sigma notation:

求和符号表示为：

f(x) = Σ [f⁽ⁿ⁾(0) / n!] xⁿ, n = 0 to ∞

For IB AQA, you must be able to write the first few terms for given functions and recognise when the expansion is valid for a certain radius of convergence.

对于 IB AQA 考试，你必须能够写出给定函数的前几项，并能判断展开式在某个收敛半径内有效。

2. Derivation from Taylor Series | 从泰勒级数推导

The Maclaurin series is a special case of the Taylor series centred at a = 0. By substituting a = 0 into the Taylor formula, the expansion simplifies significantly.

麦克劳林级数是泰勒级数在 a = 0 时的特例。把 a = 0 代入泰勒公式，展开式就大大简化了。

The Taylor series centred at a is: f(x) = Σ [f⁽ⁿ⁾(a) / n!] (x – a)ⁿ. Setting a = 0 gives the Maclaurin form. This derivation emphasises why all derivatives are evaluated at zero and why powers are simply xⁿ.

以 a 为中心的泰勒级数为：f(x) = Σ [f⁽ⁿ⁾(a) / n!] (x – a)ⁿ。令 a = 0 即得到麦克劳林形式。这个推导过程解释了为什么所有导数都在 0 处取值，以及为什么幂次都是 xⁿ。

You should be comfortable proving simple Maclaurin expansions, such as for eˣ, sin x, and cos x, by computing successive derivatives at 0 and demonstrating the pattern.

你应该能熟练地通过计算在 0 处的各阶导数并寻找规律，来证明简单的麦克劳林展开式，例如 eˣ、sin x 和 cos x。

3. Standard Maclaurin Expansions to Memorise | 必须记住的标准展开式

The IB AQA exam expects you to recall and use these standard expansions accurately:

IB AQA 考试要求你准确记忆并使用以下标准展开式：

eˣ = 1 + x + x²/2! + x³/3! + … + xⁿ/n! + …, valid for all real x.
sin x = x – x³/3! + x⁵/5! – … + (–1)ⁿ x²ⁿ⁺¹/(2n+1)! + …, valid for all real x.
cos x = 1 – x²/2! + x⁴/4! – … + (–1)ⁿ x²ⁿ/(2n)! + …, valid for all real x.
ln(1 + x) = x – x²/2 + x³/3 – … + (–1)ⁿ⁺¹ xⁿ/n + …, valid for –1 < x ≤ 1.
(1 + x)ᵏ = 1 + kx + k(k–1)x²/2! + … (binomial series), valid for |x| < 1 unless k is a non‑negative integer.

Being able to quote these instantly saves time and reduces errors when tackling more complex problems.

能够迅速写出这些展开式可以节省时间，并在处理更复杂的问题时减少错误。

4. Finding Expansions by Differentiation | 通过求导求展开式

If a function is not a standard type, you can derive its Maclaurin series by repeatedly differentiating and evaluating at 0.

如果某个函数不属于标准类型，你可以通过反复求导并在 0 处求值来推导其麦克劳林级数。

For example, to expand sec x or tan x up to the x⁴ term, you need to compute derivatives: f(0), f′(0), f″(0), f‴(0), f⁗(0). Then substitute into the formula f(0) + f′(0)x + f″(0)x²/2! + … . This method is general but can become algebraically heavy; exam questions often ask for terms up to x³ or x⁴.

例如，要展开 sec x 或 tan x 到 x⁴ 项，你需要计算 f(0), f′(0), f″(0), f‴(0), f⁗(0)，然后代入公式 f(0) + f′(0)x + f″(0)x²/2! + … 。这种方法很普遍，但代数计算可能很繁重；考题往往只要求展开到 x³ 或 x⁴ 项。

A common pitfall is forgetting to divide by factorials. Always check: term for xⁿ is f⁽ⁿ⁾(0) xⁿ / n!.

一个常见错误是忘记除以阶乘。务必检查：xⁿ 项的系数是 f⁽ⁿ⁾(0) / n!。

5. Substitution to Find New Expansions | 代入法求新展开式

Substitution is a powerful and efficient technique to find Maclaurin expansions without heavy differentiation. You replace the variable in a known series with a new expression, then simplify.

代入法是求麦克劳林展开式的一个强大而高效的方法，无需进行复杂的求导。只需把已知级数中的变量替换成一个新的表达式，再化简即可。

For example, to expand eˣ², substitute x² for x in the expansion of eˣ: eˣ² = 1 + (x²) + (x²)²/2! + … = 1 + x² + x⁴/2 + … . Similarly, sin(3x) uses the sin x series with x replaced by 3x, giving 3x – (3x)³/3! + … = 3x – 27x³/6 + … = 3x – (9/2)x³ + … .

例如，要展开 eˣ²，只需把 eˣ 展开式中的 x 替换为 x²：eˣ² = 1 + (x²) + (x²)²/2! + … = 1 + x² + x⁴/2 + … 。类似地，sin(3x) 利用 sin x 的展开式，把 x 换成 3x，得到 3x – (3x)³/3! + … = 3x – 27x³/6 + … = 3x – (9/2)x³ + … 。

Always consider the validity: if the original series converges for |x| < R, then after substitution x → g(x), the new series converges for |g(x)| < R.

务必考虑有效性：如果原级数在 |x| < R 时收敛，那么经过 x → g(x) 的代入后，新级数在 |g(x)| < R 时收敛。

6. Multiplication and Division of Series | 级数的乘除法

Exam questions sometimes ask for the Maclaurin series of a product like eˣ sin x, or a quotient like tan x. You can multiply or divide the series expansions term‑by‑term.

考题有时会要求求乘积的麦克劳林级数，如 eˣ sin x，或商的级数如 tan x。你可以对级数展开式进行逐项乘除。

For multiplication, multiply the two series as if they were polynomials, collecting terms up to the required power. For eˣ sin x up to x³: (1 + x + x²/2 + x³/6)(x – x³/6) = x + x² + (1/2 – 1/6)x³ + … = x + x² + x³/3 + … .

乘法运算时，把两个级数当成多项式相乘，累加至所需的幂次。例如 eˣ sin x 至 x³：(1 + x + x²/2 + x³/6)(x – x³/6) = x + x² + (1/2 – 1/6)x³ + … = x + x² + x³/3 + … 。

Division can be done by writing the quotient as a product with the reciprocal series, or by using polynomial long division. For example, tan x = sin x / cos x, expand both and divide. This requires careful algebraic manipulation and is a higher‑order exam skill.

除法可以通过将商写成乘以倒数级数的形式，或使用多项式长除法来完成。例如，tan x = sin x / cos x，展开两者后再相除。这需要仔细的代数运算，是较高层次的考试技能。

7. Approximating Functions | 函数近似计算

One of the main uses of Maclaurin series is to approximate function values. By truncating the series after a few terms, you obtain a polynomial approximation that is accurate near x = 0.

麦克劳林级数的主要用途之一是近似计算函数值。把级数截断为几项后，就得到一个在 x = 0 附近很精确的多项式近似。

For instance, to estimate e⁰·¹, use eˣ ≈ 1 + x + x²/2! + x³/3!. Plugging x = 0.1 gives 1 + 0.1 + 0.005 + 0.0001667 = 1.1051667, which is very close to the true value. You must be able to state the order of the approximation used.

例如，要估算 e⁰·¹，可使用 eˣ ≈ 1 + x + x²/2! + x³/3!。代入 x = 0.1 得到 1 + 0.1 + 0.005 + 0.0001667 = 1.1051667，这与真实值非常接近。你必须能够说出所使用的近似阶数。

Examiners often ask: ‘Find the Maclaurin polynomial of degree n’ – meaning you include terms up to xⁿ.

考官常常会问：“求 n 次麦克劳林多项式”——这意味着包含直到 xⁿ 的所有项。

8. Error Bounds using Lagrange Remainder | 拉格朗日余项误差界限

The Lagrange remainder form gives a rigorous bound on the error when truncating a Maclaurin series. For a series truncated at the n‑th term, the error Rₙ(x) satisfies:

拉格朗日余项形式对麦克劳林级数截断后的误差给出了严格的界限。对于在第 n 项截断的级数，误差 Rₙ(x) 满足：

Rₙ(x) = f⁽ⁿ⁺¹⁾(c) xⁿ⁺¹ / (n+1)!

for some c between 0 and x. You can bound the error by finding the maximum value of |f⁽ⁿ⁺¹⁾(c)| on that interval.

其中 c 是介于 0 与 x 之间的某个值。你可以通过求出 |f⁽ⁿ⁺¹⁾(c)| 在该区间上的最大值来界定误差。

Typical exam question: ‘Use the Maclaurin series for sin x up to x³ to estimate sin 0.2, and find an upper bound for the error.’ You would compute the 4th derivative of sin x (which is sin x, maximum magnitude 1) and substitute into the formula.

典型的考题是：“利用 sin x 的麦克劳林级数到 x³ 来估计 sin 0.2，并求误差的上界。”你需要计算 sin x 的 4 阶导数（即 sin x，最大绝对值为 1），然后代入公式。

9. Maclaurin Series for Composite Functions | 复合函数的麦克劳林级数

For composite functions like f(g(x)) when g(0) = 0, you can first expand g(x) and then substitute into the expansion of f(x). This combines substitution and multiplication techniques.

对于像 f(g(x)) 这样满足 g(0) = 0 的复合函数，可以先展开 g(x)，再代入 f(x) 的展开式。这结合了代入和乘法技巧。

Example: ln(cos x). Write cos x = 1 – x²/2 + x⁴/24 – …, so ln(cos x) = ln(1 + u) with u = – x²/2 + x⁴/24 – … . Then expand ln(1+u) = u – u²/2 + u³/3 – … and simplify, only keeping terms up to the desired power. This yields ln(cos x) = –x²/2 – x⁴/12 + … .

例如：ln(cos x)。写出 cos x = 1 – x²/2 + x⁴/24 – …，于是 ln(cos x) = ln(1 + u)，其中 u = – x²/2 + x⁴/24 – … 。然后展开 ln(1+u) = u – u²/2 + u³/3 – … ，化简后只保留到所需幂次。这样可得到 ln(cos x) = –x²/2 – x⁴/12 + … 。

Care is needed: the substitution is valid only when |u| < 1, which holds for small x.

需要注意：只有当 |u| < 1 时替换才有效，而这在 x 很小时是成立的。

10. Maclaurin Series for Inverse Trigonometric Functions | 反三角函数的麦克劳林展开

Expansions for arctan x and arcsin x can be derived by integrating known series or by binomial expansion. These appear regularly in IB AQA exams.

arctan x 和 arcsin x 的展开式可以通过对已知级数积分或二项式展开来推导。它们在 IB AQA 考试中经常出现。

The derivative of arctan x is 1/(1+x²) = 1 – x² + x⁴ – … (geometric series for |x|<1). Integrating term‑by‑term gives arctan x = x – x³/3 + x⁵/5 – ... + C, and using arctan 0 = 0 gives C=0. So arctan x = Σ (–1)ⁿ x²ⁿ⁺¹/(2n+1).

arctan x 的导数是 1/(1+x²) = 1 – x² + x⁴ – … （|x|<1 时的几何级数）。逐项积分得到 arctan x = x – x³/3 + x⁵/5 – ... + C，再由 arctan 0 = 0 得 C=0。因此 arctan x = Σ (–1)ⁿ x²ⁿ⁺¹/(2n+1)。

Similarly, arcsin x can be found using (1 – x²)⁻½ = 1 + (1/2)x² + (1·3)/(2·4) x⁴ + …, then integrating. This is a common source of 5–6 mark questions.

类似地，arcsin x 可利用 (1 – x²)⁻½ = 1 + (1/2)x² + (1·3)/(2·4) x⁴ + … 再积分得到。这是常见的 5–6 分的题目来源。

11. Using Maclaurin Series to Evaluate Limits | 运用麦克劳林级数求极限

Maclaurin series are an elegant tool for evaluating limits of the form 0/0. By expanding numerator and denominator as power series, the limit often reduces to the ratio of the leading coefficients.

麦克劳林级数是求解 0/0 型极限的优雅工具。把分子和分母都展开成幂级数后，极限通常化为首项系数之比。

Example: lim (x → 0) [sin x – x] / x³. Write sin x = x – x³/6 + x⁵/120 – …, so sin x – x = –x³/6 + x⁵/120 – … . Divide by x³: [–1/6 + x²/120 – …] → –1/6 as x → 0. This avoids multiple applications of L’Hôpital’s rule.

例如：lim (x → 0) [sin x – x] / x³。写出 sin x = x – x³/6 + x⁵/120 – …，所以 sin x – x = –x³/6 + x⁵/120 – … 。除以 x³得：[–1/6 + x²/120 – …] → –1/6（当 x → 0 时）。这样避免了多次应用洛必达法则。

Examiner tip: always expand to at least one term beyond the point where the leading parts cancel, to avoid indeterminate forms.

考官提示：务必展开到至少比首项抵消部分多一项，以免出现不定式。

12. Exam Tips and Common Pitfalls | 考试技巧与常见错误

Success in Maclaurin series questions requires both conceptual understanding and careful algebraic execution. Here are key exam tips:

要在麦克劳林级数题目中取得成功，既需要概念理解，也需要细致的代数运算。以下是一些重要的考试技巧：

Radius of convergence: Always state or check the validity interval, especially for ln(1+x) and binomial series. Omitting the condition can lose marks.
收敛半径： 始终声明或检查有效性区间，尤其是 ln(1+x) 和二项式级数。遗漏条件会失分。
Factorial errors: Make a clear table of derivatives and corresponding factorial division. f⁽ⁿ⁾(0)/n! not f⁽ⁿ⁾(0).
阶乘错误： 制作清晰的导数与对应阶乘除法表格。系数是 f⁽ⁿ⁾(0)/n!，而不是 f⁽ⁿ⁾(0)。
Keep enough terms: When asked ‘up to x⁴’, include all terms up to that power; do not miss odd or even powers accidentally.
保留足够的项： 当题目要求“到 x⁴”，应包括直到该幂次的所有项；不要意外遗漏奇数次或偶数次幂。
Substitution vs. differentiation: Substitution is faster and less error‑prone when a known series exists. Always check if the function can be rewritten into a standard form.
代入法与求导法： 当存在已知级数时，代入法更快且更不易出错。始终检查函数是否能转化为标准形式。
Error bound: Be ready to use Lagrange remainder – it’s a specific skill that distinguishes top performers.
误差界限： 做好使用拉格朗日余项的准备——这是区分高分考生的特定技能。

Consistent practice with past paper questions will build speed and confidence.

通过持续的真题练习，可以提高解题速度和信心。

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Maclaurin Series: Key Exam Points | 麦克劳林展开考点精讲