📚 Maclaurin Series Expansion | 麦克劳林展开 考点精讲
Maclaurin series is a fundamental topic in AS Mathematics, allowing us to express functions as infinite polynomials. Mastering this technique is essential for approximations, evaluating limits, and solving differential equations. This guide highlights the key exam points.
麦克劳林级数是AS数学中的一个基础课题，它允许我们将函数表示为无穷多项式。掌握这一技巧对于求近似值、计算极限和求解微分方程至关重要。本指南将梳理核心考点。
1. Definition of Maclaurin Series | 麦克劳林级数的定义
The Maclaurin series of a function f(x) is given by f(x) = f(0) + f'(0)x + f”(0)x²/2! + f”'(0)x³/3! + … + f⁽ⁿ⁾(0)xⁿ/n! + … . It is a Taylor series centred at x = 0. The series converges to f(x) for all x within its interval of convergence.
函数 f(x) 的麦克劳林级数为 f(x) = f(0) + f'(0)x + f”(0)x²/2! + f”'(0)x³/3! + … + f⁽ⁿ⁾(0)xⁿ/n! + … 。它是以 x=0 为中心的泰勒级数。在收敛区间内，该级数收敛于 f(x)。
2. Deriving the Maclaurin Series for eˣ, sin x and cos x | 推导 eˣ、sin x 和 cos x 的麦克劳林级数
For f(x)=eˣ, all derivatives are eˣ, so f⁽ⁿ⁾(0)=1. This yields eˣ = 1 + x + x²/2! + x³/3! + x⁴/4! + … . For sin x, the derivatives cycle through cos x, –sin x, –cos x, sin x; at 0 they give 0, 1, 0, –1, … Hence sin x = x – x³/3! + x⁵/5! – x⁷/7! + … . For cos x, the values are 1, 0, –1, 0, … so cos x = 1 – x²/2! + x⁴/4! – x⁶/6! + … .
对 f(x)=eˣ，所有阶导数均为 eˣ，故 f⁽ⁿ⁾(0)=1，得 eˣ = 1 + x + x²/2! + x³/3! + x⁴/4! + … 。sin x 的导数依次为 cos x、–sin x、–cos x、sin x，在 0 处取值为 0, 1, 0, –1,…，因此 sin x = x – x³/3! + x⁵/5! – x⁷/7! + … 。cos x 的相应值为 1, 0, –1, 0,…，故 cos x = 1 – x²/2! + x⁴/4! – x⁶/6! + … 。
3. Standard Maclaurin Series to Memorise | 需要记忆的标准麦克劳林展开式
The table below summarises the essential Maclaurin expansions you must know for the exam. Recognising these patterns allows quick substitutions and coefficient comparisons.
下表总结了考试中必须掌握的关键麦克劳林展开式。熟记这些形式有助于快速进行代换和系数对比。
| Function | Maclaurin Series (first few terms) | Convergence |
|---|---|---|
| eˣ | 1 + x + x²/2! + x³/3! + x⁴/4! + … | 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 (all x if n∈ℕ) |
4. Using Substitution to Find Series | 利用代换求级数展开
By substituting expressions into known series, you can expand composite functions. For example, replacing x with 2x in the eˣ series gives e²ˣ = 1 + 2x + (2x)²/2! + (2x)³/3! + … = 1 + 2x + 2x² + (4/3)x³ + … . Similarly, sin(3x) = 3x – (3x)³/3! + … = 3x – (9/2)x³ + … . Always simplify each term after substitution.
将已知展开式中的 x 代换为其他表达式，即可展开复合函数。例如，在 eˣ 的展开式中将 x 换为 2x，得到 e²ˣ = 1 + 2x + (2x)²/2! + (2x)³/3! + … = 1 + 2x + 2x² + (4/3)x³ + … 。同理，sin(3x) = 3x – (3x)³/3! + … = 3x – (9/2)x³ + … 。代换后务必将每项化简。
5. Series for ln(1+x) and (1+x)ⁿ | ln(1+x) 与 (1+x)ⁿ 的级数展开
The Maclaurin series for ln(1+x) is x – x²/2 + x³/3 – x⁴/4 + … , valid only for –1 < x ≤ 1. The binomial expansion (1+x)ⁿ = 1 + nx + n(n–1)x²/2! + n(n–1)(n–2)x³/3! + … converges for |x| < 1, unless n is a positive integer, in which case the series terminates and is valid for all x.
ln(1+x) 的麦克劳林展开式为 x – x²/2 + x³/3 – x⁴/4 + … ，仅在 –1 < x ≤ 1 内成立。二项式展开 (1+x)ⁿ = 1 + nx + n(n–1)x²/2! + n(n–1)(n–2)x³/3! + … 当 |x| < 1 时收敛；若 n 为正整数，则级数有限且对所有 x 成立。
6. Multiplication and Division of Series | 级数的乘法与除法
To find the series of a product, such as eˣ sin x, multiply the two series term‑by‑term and collect like powers up to the desired order. For division, write (denominator) × (unknown series) = numerator, then equate coefficients of x⁰, x¹, x², … to find the unknown series. Exam questions frequently ask for expansions up to x³.
求乘积如 eˣ sin x 的级数时，可将两展开式逐项相乘，合并同次幂至所需阶数。对于除法，可设 (分母) × (待求级数) = 分子，然后比较 x⁰, x¹, x², … 的系数，解得待求级数。考试中常要求展开至 x³ 项。
7. Approximation and Error Estimation | 近似计算与误差估计
Truncating a Maclaurin series after a few terms gives a polynomial approximation. For instance, sin 0.1 ≈ 0.1 – (0.1)³/6 = 0.0998333… (true value ≈ 0.0998334). The error can be bounded by the absolute value of the first omitted term. For AS level, you typically do not need the full Lagrange remainder, but should be able to state that the error is less than the next term’s magnitude.
将麦克劳林级数截断前几项可得多项式近似。例如 sin 0.1 ≈ 0.1 – (0.1)³/6 = 0.0998333…（真实值约为 0.0998334）。误差可用第一个略去项的绝对值进行界定。在AS阶段，通常不需要完整的拉格朗日余项，但应能说明误差小于下一项的绝对值。
8. Interval of Convergence | 收敛区间
Every Maclaurin series has a specific interval of convergence. For eˣ, sin x and cos x, the radius is infinite; the series converge for all real x. For ln(1+x), the interval is –1 < x ≤ 1. For (1+x)ⁿ with non‑integer n, the series converges for |x| < 1. Always state the valid range when writing a series in an exam answer.
每条麦克劳林级数都有特定的收敛区间。eˣ、sin x 和 cos x 的收敛半径为无穷大，对所有实数 x 收敛。ln(1+x) 的收敛区间为 –1 < x ≤ 1。当 n 非整数时，(1+x)ⁿ 的级数仅在 |x| < 1 收敛。考试作答时务必注明有效范围。
9. Integration and Differentiation of Series | 级数的逐项积分与微分
Within the interval of convergence, you can integrate or differentiate a Maclaurin series term‑by‑term. This technique is particularly useful for obtaining series of functions like arctan x: starting from 1/(1+x²) = 1 – x² + x⁴ – x⁶ + … for |x|<1, integrate to get arctan x = x – x³/3 + x⁵/5 – x⁷/7 + … .
在收敛区间内，可对麦克劳林级数逐项积分或微分。这一技巧对于求 arctan x 等函数的展开尤为有用：由 1/(1+x²) = 1 – x² + x⁴ – x⁶ + … (|x|<1) 积分，得到 arctan x = x – x³/3 + x⁵/5 – x⁷/7 + … 。
10. Applications to Limits and Differential Equations | 极限与微分方程应用
Maclaurin series can simplify limits that would otherwise be indeterminate. For example, lim(x→0) (sin x – x)/x³: expand sin x = x – x³/6 + …, then (x – x³/6 + … – x)/x³ = –1/6. They are also used to find power series solutions to simple differential equations, approximating the unknown function near x = 0.
麦克劳林级数可简化原本呈不定式的极限。例如 lim(x→0) (sin x – x)/x³：展开 sin x = x – x³/6 + …，代入得 (x – x³/6 + … – x)/x³ = –1/6。此外，还可用于求简单微分方程的幂级数解，从而在 x=0 附近近似未知函数。
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