Maclaurin's Series Expansion | 麦克劳林级数展开

📚 Maclaurin’s Series Expansion | 麦克劳林级数展开

The Maclaurin’s series is a powerful tool in A-Level mathematics that allows us to express a smooth function as an infinite polynomial. This expansion makes complex functions easier to differentiate, integrate, and approximate numerically.

麦克劳林级数是 A-Level 数学中的一个强大工具，它允许我们将一个光滑函数表示为无穷多项式。这种展开使复杂函数的微分、积分和数值近似变得更加简单。

1. Definition of Maclaurin’s Series | 麦克劳林级数的定义

For a function f(x) that is infinitely differentiable at x = 0, the Maclaurin series is defined as:

对于在 x = 0 处可无限次求导的函数 f(x)，麦克劳林级数定义为：

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

This infinite series reconstructs the original function from its value and derivatives at x = 0.

这个无穷级数通过函数在 x = 0 处的取值及其各阶导数来重构原函数。

In more compact sigma notation, we write:

用更简洁的西格玛记号，我们写为：

f(x) = Σₙ₌₀^∞ (f⁽ⁿ⁾(0)/n!) xⁿ

where f⁽ⁿ⁾(0) denotes the n-th derivative evaluated at x = 0.

其中 f⁽ⁿ⁾(0) 表示在 x = 0 处求值的 n 阶导数。

2. Relationship with Taylor Series | 与泰勒级数的关系

The Maclaurin series is a special case of the Taylor series centered at a = 0. The general Taylor series about x = a is:

麦克劳林级数是泰勒级数在中心 a = 0 时的特例。一般的泰勒级数在 x = a 处展开为：

f(x) = Σₙ₌₀^∞ (f⁽ⁿ⁾(a)/n!) (x − a)ⁿ

When we set a = 0, the Taylor series reduces to the Maclaurin series. This means every result we derive for Maclaurin can be generalised to Taylor expansions about other points.

当我们令 a = 0 时，泰勒级数就退化为麦克劳林级数。这意味着我们为麦克劳林推导的每个结果都可以推广到其他点的泰勒展开。

In AQA A-Level Mathematics, students are expected to know the Maclaurin series for standard functions and to use substitution to obtain expansions of related functions.

在 AQA A-Level 数学中，学生需要掌握标准函数的麦克劳林级数，并会通过代换得到相关函数的展开式。

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

The following series are essential for the exam. They are given in the AQA formula booklet, but understanding their structure helps avoid errors.

以下级数对考试至关重要。它们会在 AQA 公式册中给出，但理解其结构有助于避免错误。

eˣ = 1 + x + x²/2! + x³/3! + … for all real x

eˣ = 1 + x + x²/2! + x³/3! + … 对所有实数 x 成立
sin x = x − x³/3! + x⁵/5! − x⁷/7! + … for all real x

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! + … for all real x

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 + … for −1 < x ≤ 1

ln(1 + x) = x − x²/2 + x³/3 − x⁴/4 + … 对 −1 < x ≤ 1 成立
(1 + x)ⁿ = 1 + nx + n(n−1)x²/2! + n(n−1)(n−2)x³/3! + … for |x| < 1

(1 + x)ⁿ = 1 + nx + n(n−1)x²/2! + n(n−1)(n−2)x³/3! + … 对 |x| < 1 成立

Notice that eˣ, sin x and cos x have infinite radius of convergence, while ln(1+x) and (1+x)ⁿ are only valid on a restricted interval.

注意 eˣ、sin x 和 cos x 的收敛半径是无限的，而 ln(1+x) 和 (1+x)ⁿ 仅在有限区间内有效。

4. Deriving Maclaurin Series Directly | 直接推导麦克劳林级数

When a standard expansion is not available, we can derive the series from first principles using the definition.

当标准展开不适用时，我们可以利用定义从基本原理推导级数。

Example: Find the Maclaurin series up to x³ for f(x) = e^(2x).

例如：求 f(x) = e^(2x) 的麦克劳林级数，展开至 x³ 项。

First compute the derivatives at x = 0:

首先计算在 x = 0 处的各阶导数：

f(0) = e⁰ = 1

f(0) = e⁰ = 1
f′(x) = 2e^(2x) → f′(0) = 2

f′(x) = 2e^(2x) → f′(0) = 2
f″(x) = 4e^(2x) → f″(0) = 4

f″(x) = 4e^(2x) → f″(0) = 4
f‴(x) = 8e^(2x) → f‴(0) = 8

f‴(x) = 8e^(2x) → f‴(0) = 8

Substituting into the definition:

代入定义：

e^(2x) = 1 + 2x + (4/2!)x² + (8/3!)x³ + … = 1 + 2x + 2x² + (4/3)x³ + …

This matches substituting 2x into the standard series for eˣ, confirming the validity of substitution techniques.

这与将 2x 代入 eˣ 的标准级数得到的结果一致，证实了代换技术的正确性。

5. Using Substitution in Known Series | 在已知级数中使用代换

A faster method is to replace x in a standard expansion with an expression such as 2x, −x² or 3x. This works when the series converges for the new value.

一种更快的方法是将标准展开中的 x 替换为表达式，如 2x、−x² 或 3x。只要新值在收敛区间内，这种方法就有效。

Example: Find the first three non-zero terms of sin(x²).

例如：求 sin(x²) 的前三个非零项。

Use sin u = u − u³/3! + u⁵/5! − …, then set u = x²:

使用 sin u = u − u³/3! + u⁵/5! − …，然后令 u = x²：

sin(x²) = x² − (x²)³/6 + (x²)⁵/120 − … = x² − x⁶/6 + x¹⁰/120 − …

Notice that powers of x jump by 4, not by 2, because the substitution squares the variable.

注意 x 的幂次以 4 递增，而不是 2，因为代换使变量变为平方。

For AQA, common substitutions include e^(−x), cos(2x), ln(1+3x), and (1−x)⁻¹.

对于 AQA，常见的代换包括 e^(−x)、cos(2x)、ln(1+3x) 和 (1−x)⁻¹。

6. The Binomial Series as a Special Maclaurin Expansion | 二项式级数作为麦克劳林展开的特例

The expansion (1 + x)ⁿ is itself a Maclaurin series. When n is a positive integer, the series terminates and becomes the familiar binomial theorem.

(1 + x)ⁿ 的展开本身就是一个麦克劳林级数。当 n 是正整数时，级数终止并成为熟悉的二项式定理。

When n is negative or fractional, the series is infinite and converges only for |x| < 1.

当 n 是负数或分数时，级数是无穷的，并且仅在 |x| < 1 时收敛。

Example: Expand (1 + x)⁻² up to the x³ term.

例如：将 (1 + x)⁻² 展开到 x³ 项。

(1 + x)⁻² = 1 + (−2)x + (−2)(−3)/2! x² + (−2)(−3)(−4)/3! x³ + …

Simplifying each coefficient:

化简每一项的系数：

1 − 2x + 3x² − 4x³ + …

This pattern is useful for rational functions that can be rewritten as (1 + expression)ⁿ.

这个模式对于可以改写成 (1 + 表达式)ⁿ 的有理函数很有用。

7. Validity and Convergence | 有效性和收敛性

Each Maclaurin series is only meaningful if it converges to the original function. For AQA, you need to know the interval of convergence for the standard series.

每个麦克劳林级数只有在其收敛到原函数时才有意义。对于 AQA，你需要知道标准级数的收敛区间。

Function	Series	Convergence
eˣ	Σ xⁿ/n!	all x
sin x	Σ (−1)ⁿ x²ⁿ⁺¹/(2n+1)!	all x
cos x	Σ (−1)ⁿ x²ⁿ/(2n)!	all x
ln(1+x)	Σ (−1)ⁿ⁺¹ xⁿ/n	−1 < x ≤ 1
(1+x)ⁿ	Σ C(n,k) xᵏ	\|x\| < 1

The endpoints x = ±1 need to be checked separately; for example, ln(1+x) converges at x = 1 but diverges at x = −1.

端点 x = ±1 需要单独检验；例如，ln(1+x) 在 x = 1 收敛，但在 x = −1 发散。

8. Small Angle Approximations | 小角度近似

Truncating the Maclaurin series for sin x, cos x and tan x gives the well-known small angle approximations.

截断 sin x、cos x 和 tan x 的麦克劳林级数就得到著名的小角度近似。

sin x ≈ x for small x

sin x ≈ x 对较小的 x
cos x ≈ 1 − x²/2 for small x

cos x ≈ 1 − x²/2 对较小的 x
tan x ≈ x for small x

tan x ≈ x 对较小的 x

These approximations are used in physics and engineering. In A-Level maths, you may be asked to find a series to a given number of terms and then evaluate it for a small x.

这些近似用于物理和工程。在 A-Level 数学中，你可能会被要求找到指定项数的级数，然后用一个较小的 x 来求值。

Example: Use the first two non-zero terms of cos x to estimate cos(0.1×π).

例如：使用 cos x 的前两个非零项来估算 cos(0.1×π)。

cos(0.31416) ≈ 1 − (0.31416)²/2 = 1 − 0.04935 = 0.95065

The actual value is 0.95106, giving a very small error.

实际值为 0.95106，误差非常小。

9. Differentiating and Integrating Series | 级数的微分与积分

Because a Maclaurin series is a polynomial-like expression, it can be differentiated or integrated term-by-term within its interval of convergence.

由于麦克劳林级数是类似多项式的表达式，在其收敛区间内可以逐项微分或积分。

Example: Starting from the series for eˣ, find the series for xeˣ by multiplication, or for e^(x²) by substitution.

例如：从 eˣ 的级数出发，通过乘法求 xeˣ 的级数，或通过代换求 e^(x²) 的级数。

Integrating the series for 1/(1+x) = 1 − x + x² − x³ + … gives the series for ln(1+x):

对 1/(1+x) = 1 − x + x² − x³ + … 的级数积分得到 ln(1+x) 的级数：

∫ 1/(1+x) dx = x − x²/2 + x³/3 − x⁴/4 + … = ln(1+x)

This is a nice derivation of a standard result using integration of a geometric series.

这是通过对几何级数积分来推导一个标准结果的漂亮方法。

10. Products of Series | 级数的乘积

To find the Maclaurin series of a product like eˣ sin x, multiply the truncated series and collect like powers of x.

要求乘积如 eˣ sin x 的麦克劳林级数，将截断后的级数相乘并合并同类项。

Example: Find the series up to x³ for eˣ sin x.

例如：求 eˣ sin x 的级数，展开至 x³ 项。

eˣ sin x = (1 + x + x²/2 + x³/6)(x − x³/6) + …

Multiply and keep terms up to x³:

相乘并保留到 x³ 项：

1 × x = x

1 × x = x
x × x = x²

x × x = x²
x²/2 × x = x³/2

x²/2 × x = x³/2
1 × (−x³/6) = −x³/6

1 × (−x³/6) = −x³/6

So eˣ sin x = x + x² + (1/2 − 1/6)x³ = x + x² + (1/3)x³ + …

所以 eˣ sin x = x + x² + (1/2 − 1/6)x³ = x + x² + (1/3)x³ + …

This technique appears frequently in AQA exam questions, especially in Paper 2.

这种技巧在 AQA 考试题目中经常出现，尤其是在卷二。

11. Worked AQA-Style Example | AQA 风格例题

A typical exam question: Given f(x) = ln(1+2x), find the Maclaurin series up to the x³ term, and state the range of values of x for which the expansion is valid.

典型考试题：已知 f(x) = ln(1+2x)，求麦克劳林级数至 x³ 项，并说明展开式成立的 x 取值范围。

Method 1: Substitute 2x into the standard series for ln(1+u):

方法一：将 2x 代入 ln(1+u) 的标准级数：

ln(1+2x) = 2x − (2x)²/2 + (2x)³/3 − … = 2x − 2x² + (8/3)x³ − …

The standard series converges for −1 < u ≤ 1, so we need −1 < 2x ≤ 1, giving −1/2 < x ≤ 1/2.

标准级数在 −1 < u ≤ 1 收敛，所以需要 −1 < 2x ≤ 1，得到 −1/2 < x ≤ 1/2。

Method 2: Differentiate f(x) three times and evaluate at x = 0:

方法二：对 f(x) 求导三次并在 x = 0 求值：

f(0) = ln 1 = 0

f(0) = ln 1 = 0
f′(x) = 2/(1+2x) → f′(0) = 2

f′(x) = 2/(1+2x) → f′(0) = 2
f″(x) = −4/(1+2x)² → f″(0) = −4

f″(x) = −4/(1+2x)² → f″(0) = −4
f‴(x) = 16/(1+2x)³ → f‴(0) = 16

f‴(x) = 16/(1+2x)³ → f‴(0) = 16

Then using the definition: f(x) = 0 + 2x + (−4/2)x² + (16/6)x³ = 2x − 2x² + (8/3)x³. Both methods give the same answer.

然后使用定义：f(x) = 0 + 2x + (−4/2)x² + (16/6)x³ = 2x − 2x² + (8/3)x³。两种方法得到相同结果。

12. Common Mistakes and Exam Tips | 常见错误与考试技巧

Students often make errors in the following areas. Be careful to avoid them:

学生常常在以下方面犯错。请小心避免：

Forgetting factorials in the denominator of the general term.

忘记通项分母中的阶乘。
Using the Taylor series formula but evaluating derivatives at the wrong point.

使用泰勒级数公式但在错误的点求导数值。
State the validity of the expansion incorrectly, especially for ln(1+x) and (1+x)ⁿ.

错误说明展开式的有效性，特别是 ln(1+x) 和 (1+x)ⁿ。
When multiplying series, dropping a term that could contribute to a lower power of x.

相乘级数时，漏掉可能对较低次幂有贡献的项。
Using degrees instead of radians when evaluating small angle approximations.

在小角度近似中使用度数而不是弧度。

Exam tips:

考试建议：

Always state the interval of convergence when asked.

被要求时始终说明收敛区间。
Check whether you need a specific number of non-zero terms, not just terms up to a power.

检查你需要的是特定数量的非零项，而不仅仅是到某次幂。
For ln(1+x), remember the alternating signs and the factor of 1/n, not 1/n!.

对于 ln(1+x)，记住交替的符号和 1/n 因子，而不是 1/n!。

With practice, Maclaurin expansions become a reliable source of marks in the AQA exam. Master the standard series, then perfect your substitution and multiplication techniques.

通过练习，麦克劳林展开会成为 AQA 考试中可靠的得分点。掌握标准级数，然后完善你的代换和乘法技巧。

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Maclaurin’s Series Expansion | 麦克劳林级数展开