Maclaurin Series | 麦克劳林展开

📚 Maclaurin Series | 麦克劳林展开

Maclaurin series is a powerful tool in A Level Mathematics that expresses a function as an infinite sum of terms calculated from the values of its derivatives at a single point — zero. It is a special case of the Taylor series, centred at x = 0. For Edexcel students, mastering Maclaurin expansions is essential not only for approximating functions but also for evaluating limits, solving differential equations, and understanding the behaviour of functions near the origin. This article will walk you through the key concepts, standard expansions, derivation techniques, validity conditions, and typical exam applications.

麦克劳林级数是 A Level 数学中的一个有力工具，它将一个函数表示为无穷多项的和，这些项由函数在零点处的各阶导数值计算得出。它是以 x = 0 为中心的泰勒级数的特例。对于学习 Edexcel 课程的学生来说，掌握麦克劳林展开不仅对逼近函数至关重要，还用于求极限、解微分方程以及理解函数在原点附近的行为。本文将带你梳理核心概念、标准展开式、推导方法、有效性条件以及典型的考试应用。

1. Definition and Formula | 定义与公式

The Maclaurin series for a function f(x) that is infinitely differentiable at x = 0 is given by:
f(x) = f(0) + f'(0)x + f”(0)x²/2! + f”'(0)x³/3! + … + f⁽ⁿ⁾(0)xⁿ/n! + …
In sigma notation: f(x) = Σ [f⁽ⁿ⁾(0) / n!] xⁿ, summing from n = 0 to ∞. The general term involves the nth derivative evaluated at zero, divided by n factorial, multiplied by x raised to the nth power.

一个在 x = 0 处无穷次可导的函数 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! + …
用求和符号表示为：f(x) = Σ [f⁽ⁿ⁾(0) / n!] xⁿ，求和指标 n 从 0 到 ∞。通项由函数在零点的 n 阶导数值除以 n 的阶乘，再乘以 x 的 n 次方构成。

To generate a Maclaurin expansion, you must compute the function’s value and all necessary derivatives at x = 0, then substitute into the formula. For most exam problems, you will only need the first few non‑zero terms.

要生成麦克劳林展开式，必须先计算出函数及其各阶导数在 x = 0 处的值，再代入公式。在绝大多数考试题目中，你只需要展开前几项非零项即可。

2. Standard Maclaurin Expansions | 标准麦克劳林展开式

Edexcel expects you to know and be able to use these standard expansions without deriving them from scratch each time:
eˣ = 1 + x + x²/2! + x³/3! + … valid for all x
sin x = x − x³/3! + x⁵/5! − x⁷/7! + … valid for all x
cos x = 1 − x²/2! + x⁴/4! − x⁶/6! + … valid for all x
ln(1 + x) = x − x²/2 + x³/3 − x⁴/4 + … valid for −1 < x ≤ 1 (1 + x)ⁿ = 1 + nx + n(n−1)x²/2! + n(n−1)(n−2)x³/3! + ... valid for |x| < 1 (unless n is a positive integer).

Edexcel 考试要求学生熟悉并能够直接使用以下标准展开式，而不必每次从头推导：
eˣ = 1 + x + x²/2! + x³/3! + … 对所有实数 x 成立
sin x = x − x³/3! + x⁵/5! − x⁷/7! + … 对所有实数 x 成立
cos x = 1 − x²/2! + x⁴/4! − x⁶/6! + … 对所有实数 x 成立
ln(1 + x) = x − x²/2 + x³/3 − x⁴/4 + … 成立范围为 −1 < x ≤ 1 (1 + x)ⁿ = 1 + nx + n(n−1)x²/2! + n(n−1)(n−2)x³/3! + ... 在 |x| < 1 时成立（除非 n 为正整数）。

Memorising the general patterns — alternating signs for sin and cos, factorial denominators, the logarithmic expansion lacking factorials — saves time and reduces errors. You should also recognise that the expansion of eˣ has all positive coefficients, whereas sin and cos alternate.

记忆这些通项模式——sin 和 cos 的正负交变、阶乘分母、对数展开不含阶乘——可以节约时间并减少错误。你还要能识别 eˣ 的展开系数全部为正，而 sin 和 cos 是正负交替的。

3. Deriving Expansions by Differentiation | 通过求导推导展开式

When a function is not among the standard forms, or when the exam asks you to show the derivation, you differentiate repeatedly to find f(0), f'(0), f”(0), f”'(0), etc. Each derivative is evaluated at x = 0 before proceeding to the next order. For example, to expand f(x) = sec x up to the x⁴ term, compute:
f(0) = sec 0 = 1
f'(x) = sec x tan x, f'(0) = 0
f”(x) = sec x tan² x + sec³ x, f”(0) = 1
Next, f”'(x) yields f”'(0) = 0, and f⁽⁴⁾(0) = 5. Thus sec x ≈ 1 + x²/2 + 5x⁴/24.

如果所给函数不在标准形式之内，或者题目要求展示推导过程，你便需反复求导以得到 f(0), f'(0), f”(0), f”'(0) 等。每求一阶导，便代入 x = 0 计算后再求下一阶。例如，展开 f(x) = sec x 至 x⁴ 项：先计算 f(0)=sec 0=1；f'(x)=sec x tan x，f'(0)=0；f”(x)=sec x tan²x + sec³x，f”(0)=1；接着 f”'(0)=0，而 f⁽⁴⁾(0)=5。因此 sec x ≈ 1 + x²/2 + 5x⁴/24。

Notice that only even powers appear because sec x is an even function. This serves as a useful check: an even function’s Maclaurin series contains only even powers of x, while an odd function’s series contains only odd powers.

注意到因为 sec x 是偶函数，其展开式仅含偶次幂。这是一个实用的检验方法：偶函数的麦克劳林级数只包含 x 的偶次幂，奇函数只包含奇次幂。

4. Validity and Convergence | 有效性与收敛性

Every Maclaurin series has an interval of convergence, which tells you the x‑values for which the infinite sum equals the function. The radius of convergence is often found using the ratio test. For standard expansions, validity ranges are: eˣ, sin x, cos x converge for all real x; ln(1 + x) converges for −1 < x ≤ 1; (1 + x)ⁿ converges for |x| < 1 (if n is not a positive integer). When combining series (e.g., multiplying by a polynomial), the validity remains the more restrictive of the component validities.

每个麦克劳林级数都有一个收敛区间，该区间指明了使无穷级数等于原函数的 x 取值范围。收敛半径通常用比值判别法求得。对于标准展开式，有效范围是：eˣ、sin x、cos x 对所有实数 x 收敛；ln(1 + x) 在 −1 < x ≤ 1 时收敛；(1 + x)ⁿ 若非正整数幂，则在 |x| < 1 时收敛。当对级数进行组合时（例如乘以一个多项式），其有效性取各组成部分中更严格的限制。

Edexcel questions often ask you to state the range of validity after finding an expansion. Always check whether the function is defined at x = 0 and whether any singularities restrict the convergence. For example, ln(1 + x) is undefined at x = −1, so the series cannot converge there; it does converge at x = 1, though the rate slows down.

Edexcel 试题经常要求你在求出展开式之后说明其有效范围。务必检查函数在 x = 0 处是否有定义，以及是否存在奇点限制了收敛。例如，ln(1 + x) 在 x = −1 处无定义，因此级数在那里不可能收敛；而在 x = 1 处级数收敛，虽然收敛速度很慢。

5. Composite Functions | 复合函数的展开

To expand a function like eˢⁱⁿ ˣ or ln(1 + sin x), you can substitute one standard series into another. For eˢⁱⁿ ˣ, write sin x = x − x³/6 + … and replace u in eᵘ = 1 + u + u²/2 + u³/6 + … . Then collect terms in powers of x, keeping only up to the required order. For instance, u = x − x³/6, so u² = x² − x⁴/3 + … (ignoring higher powers). Substituting gives eˢⁱⁿ ˣ ≈ 1 + (x − x³/6) + (x² − x⁴/3)/2 + x³/6 = 1 + x + x²/2 − x⁴/8 + … (up to x⁴).

对于形如 eˢⁱⁿ ˣ 或 ln(1 + sin x) 的函数，可以将一个标准级数代入另一个标准级数中进行展开。以 eˢⁱⁿ ˣ 为例，先写出 sin x = x − x³/6 + …，再令 eᵘ = 1 + u + u²/2 + u³/6 + …，用 u = x − x³/6 替换，然后按 x 的幂次整理各项，仅保留所需阶数。例如，u² = x² − x⁴/3 + …（忽略更高次），代入可得 eˢⁱⁿ ˣ ≈ 1 + (x − x³/6) + (x² − x⁴/3)/2 + x³/6 = 1 + x + x²/2 − x⁴/8 + …（至 x⁴ 项）。

Always check that the substitution stays within the region of convergence of the outer series. If the inner series produces values outside the valid range of the outer series, the composite expansion may be invalid for some x.

务必确保代换后的值仍落在外层级数的收敛区域内。若内层级数的值超出了外层级数的有效范围，复合展开式对某些 x 可能不再成立。

6. Products and Quotients of Series | 级数的乘积与商

When you need to expand the product of two functions, multiply their known Maclaurin series and collect like powers. For example, to expand eˣ sin x up to x³, use eˣ = 1 + x + x²/2 + x³/6 + … and sin x = x − x³/6 + … . Their product up to x³ is:
x + x² + (1/2 − 1/6)x³ = x + x² + x³/3, giving eˣ sin x ≈ x + x² + x³/3.

当需要展开两个函数的乘积时，可将它们已知的麦克劳林级数相乘，然后合并同次幂项。例如，展开 eˣ sin x 至 x³，利用 eˣ = 1 + x + x²/2 + x³/6 + … 和 sin x = x − x³/6 + …，它们的乘积至 x³ 为：x + x² + (1/2 − 1/6)x³ = x + x² + x³/3，因此 eˣ sin x ≈ x + x² + x³/3。

To obtain a quotient like tan x = sin x / cos x, write sin x and cos x as series, then use algebraic division or equate coefficients by writing tan x = a₀ + a₁x + a₂x² + a₃x³ + … and cross‑multiplying with the cosine series. You will obtain a system of equations for the coefficients, yielding tan x = x + x³/3 + 2x⁵/15 + …

若要展开像 tan x = sin x / cos x 这样的商，可将 sin x 和 cos x 写成级数，然后通过长除法或待定系数法求解。设 tan x = a₀ + a₁x + a₂x² + a₃x³ + …，与 cos x 的级数交叉相乘，得到一个关于系数的方程组，最终解得 tan x = x + x³/3 + 2x⁵/15 + …

7. Integration and Differentiation of Series | 级数的积分与求导

A powerful technique within the radius of convergence is to integrate or differentiate a known series term‑by‑term. If you know that 1/(1 − x) = 1 + x + x² + x³ + … for |x| < 1, then integrating gives −ln(1 − x) = x + x²/2 + x³/3 + x⁴/4 + ... . Substituting −x for x yields the standard ln(1 + x) series. Similarly, differentiating the series for sin x confirms the series for cos x.

在收敛半径内，可以对已知级数进行逐项积分或逐项求导，这是一个强有力的技巧。已知 1/(1 − x) = 1 + x + x² + x³ + … (|x| < 1)，积分后得到 −ln(1 − x) = x + x²/2 + x³/3 + x⁴/4 + ...。将 x 换成 −x 便得到标准的 ln(1 + x) 级数。同理，对 sin x 的级数求导可以验证 cos x 的级数。

Term‑by‑term operations are especially useful when a function is defined by an integral, such as Si(x) = ∫ sin t / t dt. Expanding sin t / t = 1 − t²/3! + t⁴/5! − … and integrating yields Si(x) ≈ x − x³/18 + x⁵/600 − … . Edexcel exams may ask you to derive such series using integration.

当函数由积分定义时，逐项积分尤其有用，例如 Si(x) = ∫ sin t / t dt。展开 sin t / t = 1 − t²/3! + t⁴/5! − … 然后积分，便得到 Si(x) ≈ x − x³/18 + x⁵/600 − … 。Edexcel 试题可能要求你利用积分来推导此类级数。

8. Approximating Functions and Error Estimation | 函数逼近与误差估计

A truncated Maclaurin series gives a polynomial approximation to f(x) near zero. The error in using the series up to the term in xⁿ is given by the Lagrange remainder: Rₙ(x) = f⁽ⁿ⁺¹⁾(c) xⁿ⁺¹/(n+1)! for some c between 0 and x. In practice, for small x you can estimate the maximum error by bounding the next derivative. For instance, approximating e⁰·¹ using the first four terms yields an error less than about 2.8 × 10⁻⁵ because the next term is (0.1)⁴/4! with the exponential derivative bounded by e⁰·¹.

截断的麦克劳林级数给出了 f(x) 在零点附近的多项式逼近。使用到 xⁿ 项为止的级数所产生的误差可用拉格朗日余项表示：Rₙ(x) = f⁽ⁿ⁺¹⁾(c) xⁿ⁺¹/(n+1)!，其中 c 为 0 与 x 之间的某值。实际应用中，对于较小的 x，你可以通过放大下一阶导数来估计最大误差。例如，用前四项近似 e⁰·¹ 时，误差小于约 2.8 × 10⁻⁵，因为下一项为 (0.1)⁴/4!，且指数函数的导数被 e⁰·¹ 所限定。

Exam questions often ask you to approximate an integral or evaluate a function to a specified accuracy. You should determine how many terms are needed by ensuring the next term is smaller than the required tolerance. This links directly to the concept of convergence and to real‑world precision.

考试题常要求你近似求积分或按给定精度计算函数值。你应当通过确保下一项小于所需容差来确定需要多少项。这直接关联到收敛性的概念以及实际应用中的精度要求。

9. Evaluating Limits using Maclaurin Series | 利用麦克劳林展开求极限

Maclaurin series turn complicated limit expressions into ratios of polynomials, allowing you to cancel common factors and evaluate the limit directly. For instance, lim(x→0) (eˣ − 1 − x) / x² can be found by substituting eˣ = 1 + x + x²/2 + x³/6 + …, giving (x²/2 + x³/6 + …)/x² = 1/2 + x/6 + … → 1/2 as x → 0. This method is especially useful when L’Hôpital’s Rule would be lengthy or when indeterminate forms involve higher order terms.

麦克劳林级数能将复杂的极限表达式转化为多项式之比，使你能够约去公因子并直接求出极限。例如，lim(x→0) (eˣ − 1 − x) / x²，代入 eˣ = 1 + x + x²/2 + x³/6 + …，得到 (x²/2 + x³/6 + …)/x² = 1/2 + x/6 + … → 1/2（当 x → 0）。当洛必达法则计算冗长或不定式中涉及高阶项时，这种方法尤为实用。

For limits such as lim(x→0) (sin x − x) / x³, expand sin x = x − x³/6 + x⁵/120 − … so the numerator becomes −x³/6 + x⁵/120 − …; dividing by x³ gives −1/6 + x²/120 − … → −1/6. This is far quicker than triple application of L’Hôpital’s Rule.

对于如 lim(x→0) (sin x − x) / x³ 的极限，展开 sin x = x − x³/6 + x⁵/120 − …，分子即为 −x³/6 + x⁵/120 − …；除以 x³ 得 −1/6 + x²/120 − … → −1/6。这比使用三次洛必达法则要快得多。

10. Exam Techniques and Common Pitfalls | 考试技巧与常见错误

Edexcel examiners look for method marks: always show the successive derivatives and their values at zero, even if you know the final series. Write the series with factorials in the denominator; do not prematurely evaluate them. List the first few terms clearly, specifying the order of the expansion you are keeping. When combining series, explicitly state the terms you are discarding as “higher order terms” (often written as + …). For validity, always give the interval in inequality form, such as |x| < 1, and note endpoint behaviour where relevant.

Edexcel 阅卷者看重方法分：即使你已经知道最终级数，也要展示逐步求导及其在零点的值。将级数以分母含有阶乘的形式写出，不要提前算出阶乘值。清晰地列出前几项，并指明你所保留的展开阶数。在组合级数时，明确地将丢弃的项表示为“更高阶项”（通常写作 + …）。关于有效性，始终用不等式形式给出区间，如 |x| < 1，并在相关处说明端点行为。

Common mistakes include: forgetting to divide by factorials; mis‑evaluating f(0) when the function is not defined at zero (then it cannot be expanded); applying the binomial expansion for (1+x)ⁿ when x is negative without checking the validity; confusing the alternating signs in sin and cos; and attempting to use Maclaurin expansion for functions with a vertical asymptote at x = 0, like ln x or 1/x. Remember: a Maclaurin series exists only if the function is infinitely differentiable at 0.

常见错误包括：忘记除以阶乘；当函数在零处无定义时仍错误地展开（此时无法展开）；在未检查有效性的情况下对负 x 使用 (1+x)ⁿ 的二项展开式；混淆 sin 和 cos 的交替符号；以及对在 x = 0 处有垂直渐近线的函数（如 ln x 或 1/x）尝试使用麦克劳林展开。请记住：只有函数在 0 处无穷次可导时，麦克劳林级数才存在。

Always verify your answer by substituting a small value (e.g., x = 0.1) into both the original function and your series to check they match to the claimed order of accuracy. This quick sanity check can catch algebraic errors.

务必用一个小数值（如 x = 0.1）代入原函数和你的级数，检验两者在声称的精度阶数下是否吻合。这样快速的合理性检查可以发现代数错误。

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Maclaurin Series | 麦克劳林展开