Maclaurin Series Revision for AQA Mathematics | AQA 数学：麦克劳林展开考点精讲

📚 Maclaurin Series Revision for AQA Mathematics | AQA 数学：麦克劳林展开考点精讲

Maclaurin series expansions let you replace tricky functions with friendly polynomials. This revision guide covers everything from the basic formula to error bounds, helping you master AQA exam questions on series expansions of e^x, sin x, cos x, ln(1+x), (1+x)ⁿ and beyond.

麦克劳林展开让你用简单的多项式替代复杂的函数。本篇考点精讲涵盖从基本公式到误差界的方方面面，助你攻克 AQA 考试中关于 e^x、sin x、cos x、ln(1+x)、(1+x)ⁿ 等函数的级数展开题。

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 f(x) centred at x = 0. It writes the function as an infinite sum of powers of x, with coefficients built from the derivatives of f at zero.

麦克劳林级数是函数 f(x) 在 x = 0 处的泰勒级数展开。它将函数写成 x 的幂次项的无限和，系数由 f 在零点的各阶导数构建而成。

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

The notation f⁽ⁿ⁾(0) means the n‑th derivative of f evaluated at 0, and n! is n factorial. If you can differentiate f repeatedly at 0, you can build its Maclaurin series.

记号 f⁽ⁿ⁾(0) 表示 f 在 0 处的 n 阶导数，n! 是 n 的阶乘。只要你能在 0 点反复求导，就能写出函数的麦克劳林级数。

2. Step‑by‑Step Derivation | 一步步推导展开式

To find a Maclaurin series up to a required power, follow these four steps: evaluate f(0), compute f'(0), f”(0), f”'(0) and so on until the desired order, substitute into the formula, and simplify the coefficients.

要写出指定阶数的麦克劳林展开式，按以下四步走：计算 f(0)，求出 f'(0)、f”(0)、f”'(0) 等直到所需阶数，代入通式，然后化简系数。

For example, to expand e^x up to x³: f(x)=e^x, so f(0)=1. Every derivative is also e^x, so f'(0)=1, f”(0)=1, f”'(0)=1. Plugging in gives 1 + 1·x + (1/2!)x² + (1/3!)x³ + …

例如展开 e^x 至 x³ 项：f(x)=e^x，则 f(0)=1。每一阶导数仍然是 e^x，因此 f'(0)=1, f”(0)=1, f”'(0)=1。代入得到 1 + 1·x + (1/2!)x² + (1/3!)x³ + …

Always remember to divide by the factorial of the derivative’s order – a common slip is to write x²/2 without the factorial denominator 2! .

务必牢记除以导数阶数的阶乘——常见的错误是把 x²/2 的分母写成 2 而不是 2!。

3. Must‑Know Standard Series | 必背的标准展开式

The AQA specification expects you to know these five expansions by heart, together with their ranges of validity. They form the building blocks for many exam questions.

AQA 考纲要求你熟记下面五个展开式及其有效区间。它们是大多数考题的基石。

e^x = 1 + x + x²/2! + x³/3! + … , all real x

sin x = x − x³/3! + x⁵/5! − … , all real x

cos x = 1 − x²/2! + x⁴/4! − … , 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 (or all x if n ∈ ℕ)

The series for sin x and cos x alternate in sign; ln(1+x) also alternates. The binomial expansion with a positive integer n terminates and is valid for all real x.

sin x 和 cos x 的级数正负交替；ln(1+x) 的符号也交替。当 n 为正整数时，二项式展开会终止，并且对所有实数 x 都有效。

4. Range of Validity in Detail | 收敛区间详解

The validity interval tells you where the infinite polynomial actually equals the function. For e^x, sin x and cos x, the series converge for every real number x, so you never need to worry about restrictions.

有效区间告诉你无穷多项式在什么范围内真的等于原函数。对于 e^x、sin x 和 cos x，级数对所有实数都收敛，所以完全不用担心限制条件。

However, ln(1+x) only works for −1 < x ≤ 1. The series diverges outside that range, and even at x = −1 it blows up. For (1+x)ⁿ, the condition is |x| < 1 unless n is a positive integer – then the series is a finite polynomial and holds for all x.

但 ln(1+x) 仅在 −1 < x ≤ 1 时有效。超出这个范围级数发散，在 x = −1 处甚至趋于无穷。对 (1+x)ⁿ 而言，要求 |x| < 1，除非 n 是正整数——此时展开是有限多项式，对所有 x 都成立。

A typical exam question might ask: “For what values of x is the expansion of ln(1+2x) valid?” You must solve −1 < 2x ≤ 1, giving −½ < x ≤ ½.

典型的考题会问：“ln(1+2x) 的展开式对哪些 x 有效？”你需要解 −1 < 2x ≤ 1，得到 −½ < x ≤ ½。

5. Substitution Method | 代换法求展开式

You can generate new expansions quickly by substituting a multiple of x into a known series. Replace every x in the standard series with kx, then simplify – but remember to adjust the validity interval.

你可以通过向已知级数中代入 x 的倍数来快速得到新的展开。将标准级数中的每一个 x 替换为 kx，然后化简——但别忘了调整有效区间。

For instance, to expand e^2x, start with e^x = 1 + x + x²/2! + …, then replace x by 2x: e^2x = 1 + 2x + (2x)²/2! + (2x)³/3! + … = 1 + 2x + 4x²/2! + 8x³/3! + …

例如，展开 e^2x，从 e^x = 1 + x + x²/2! + … 出发，用 2x 替换 x：e^2x = 1 + 2x + (2x)²/2! + (2x)³/3! + … = 1 + 2x + 4x²/2! + 8x³/3! + …

Similarly, sin(3x) = 3x − (3x)³/3! + (3x)⁵/5! − … = 3x − 27x³/6 + 243x⁵/120 − … All such expansions remain valid for all x, but for ln(1+3x) you must keep −1 < 3x ≤ 1, i.e. −⅓ < x ≤ ⅓.

类似地，sin(3x) = 3x − (3x)³/3! + (3x)⁵/5! − … = 3x − 27x³/6 + 243x⁵/120 − … 这些展开对所有 x 都成立，但对 ln(1+3x) 必须满足 −1 < 3x ≤ 1，即 −⅓ < x ≤ ⅓。

6. Products of Series | 级数的乘积

When a function is the product of two standard ones, multiply their series together, keeping only terms up to the required power. Work systematically so you don’t miss any cross-terms.

当函数是两个标准函数的乘积时，将它们对应的级数相乘，只保留到指定阶数的项。有条理地进行，以免遗漏交叉项。

Suppose you need e^x cos x up to x³. Write e^x = 1 + x + x²/2 + x³/6 + … and cos x = 1 − x²/2 + … (no x³ term needed because it would produce higher powers). Multiply: (1)(1) = 1; (1)(−x²/2) + (x)(x terms?) – here better to list terms: constant: 1; x term: 1·x = x; x² term: 1·(−x²/2) + (x²/2)·1 = 0; x³ term: only from e^x’s x³/6 and x·(−x²/2)? Let’s do carefully: cos x ≈ 1 − x²/2. Multiply by e^x = 1+x+x²/2+x³/6. Constant: 1. ×1: 1; x: x·1 = x; x²: (x²/2)·1 + 1·(−x²/2) = 0; x³: (x³/6)·1 + x·(−x²/2) = x³/6 − x³/2 = −x³/3. Hence e^x cos x ≈ 1 + x − x³/3. Always double-check with direct differentiation if possible.

假设你需要 e^x cos x 到 x³ 项。写出 e^x = 1 + x + x²/2 + x³/6 + … 和 cos x = 1 − x²/2 + …（不用写 x³ 项，因为会产生更高次）。相乘：常数项 1；x 项 1·x = x；x² 项：1·(−x²/2) + (x²/2)·1 = 0；x³ 项：来自 (x³/6)·1 加上 x·(−x²/2) = x³/6 − x³/2 = −x³/3。因此 e^x cos x ≈ 1 + x − x³/3。可能的话用直接求导验证一下。

7. Using Series for Approximations | 用级数求近似值

Truncating a Maclaurin series gives a polynomial approximation that is easy to evaluate. You can estimate function values like e^0.2, sin 0.1 or ln 1.05 quickly without a calculator.

截断麦克劳林级数会给出易于计算的多项式近似。你无需计算器就能快速估算 e^0.2、sin 0.1 或 ln 1.05 等函数值。

For example, approximate e^0.2 using the first three terms: 1 + 0.2 + (0.2)²/2! = 1 + 0.2 + 0.04/2 = 1.22. The true value is about 1.22140, so the error is tiny.

例如，用前三项估计 e^0.2：1 + 0.2 + (0.2)²/2! = 1 + 0.2 + 0.04/2 = 1.22。真实值约 1.22140，误差极小。

When approximating ln(1+x) for small x, say ln(1.03), use x = 0.03 in x − x²/2 + x³/3 − … The alternating signs make it easy to bound the error.

近似小 x 时的 ln(1+x) 时，比如 ln(1.03)，令 x = 0.03 代入 x − x²/2 + x³/3 − … 符号交替使得误差很容易限定。

8. Error Bounds for Alternating Series | 交替级数的误差界

If a Maclaurin series alternates in sign (like sin x, cos x, ln(1+x)), the absolute error after truncating is less than the absolute value of the first omitted term. This gives a rigorous error bound.

若麦克劳林级数正负交替（如 sin x、cos x、ln(1+x)），截断后的绝对误差小于第一个被省略项的绝对值。这给出了严格的误差界。

For instance, approximate sin 0.2 with x − x³/6. The next term is x⁵/120 = (0.2)⁵/120 ≈ 0.00032/120 ≈ 2.67 × 10⁻⁶, so the error is less than 3 × 10⁻⁶. The approximation sin 0.2 ≈ 0.2 − (0.008)/6 = 0.1986667 is extremely accurate.

例如，用 x − x³/6 近似 sin 0.2。下一项是 x⁵/120 = (0.2)⁵/120 ≈ 0.00032/120 ≈ 2.67×10⁻⁶，因此误差小于 3×10⁻⁶。近似值 sin 0.2 ≈ 0.2 − 0.008/6 = 0.1986667 非常精确。

The same logic applies to cos x: truncating after 1 − x²/2 gives an error smaller than |x⁴/4!|. For ln(1+x) near x = 0.1, stopping at x − x²/2 leaves an error less than 0.1³/3 ≈ 0.000333.

同样的逻辑适用于 cos x：在 1 − x²/2 后截断，误差小于 |x⁴/4!|。对 ln(1+x) 在 x=0.1 附近，算到 x − x²/2 时误差小于 0.1³/3 ≈ 0.000333。

9. Common Pitfalls and How to Avoid Them | 常见易错点与避坑指南

Mistake 1: forgetting to divide by the factorial. The term in x³ must be divided by 3! = 6, not 3. Always write the factorial explicitly.

错误一：忘记除以阶乘。x³ 项必须除以 3! = 6，而不是 3。务必将阶乘写清楚。

Mistake 2: ignoring the range of validity. Applying ln(1+x) at x = 2 gives a divergent series. Always state or check the allowed x interval.

错误二：忽略有效区间。在 x=2 处使用 ln(1+x) 会得到发散级数。始终给出或检查允许的 x 范围。

Mistake 3: mixing signs. The series for sin x and cos x alternate, and missing a minus sign leads to a completely wrong approximation. Double-check the sign pattern.

错误三：正负号混淆。sin x 和 cos x 的级数符号交替，漏掉一个负号将导致近似完全错误。仔细核对符号规律。

Mistake 4: using the wrong centre. Maclaurin series requires x = 0. If the centre is a different value you need the Taylor formula. Don’t shift it accidentally.

错误四：用错展开中心。麦克劳林级数要求中心为 x=0。若中心不同你需使用泰勒公式。不要不小心偏移中心。

Mistake 5: sloppy substitution. When replacing x by 2x in ln(1+x), you must also adjust the validity condition: ln(1+2x) is valid for −½ < x ≤ ½, not x > −1.

错误五：代换不严密。将 ln(1+x) 中的 x 替换为 2x 时，有效条件也必须调整：ln(1+2x) 在 −½ < x ≤ ½ 才有效，而不是 x > −1。

10. Worked Example – Combining Expansions | 综合范例——组合展开

Find the Maclaurin series of f(x) = ln(1 + sin x) up to and including the term in x³.

求 f(x) = ln(1 + sin x) 的麦克劳林级数，包含至 x³ 项。

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Maclaurin Series Revision for AQA Mathematics | AQA 数学：麦克劳林展开 考点精讲