Maclaurin Series Expansion | 麦克劳林展开考点精讲

📚 Maclaurin Series Expansion | 麦克劳林展开考点精讲

The Maclaurin series is a powerful tool for approximating functions using polynomials. In CCEA GCSE Further Mathematics, you are expected to derive and use series expansions for standard functions such as eˣ, sin x, cos x, ln(1+x) and (1+x)ⁿ. Understanding the formula and its applications will significantly boost your exam performance.

麦克劳林级数是用多项式逼近函数的重要工具。在 CCEA GCSE 进阶数学中，你需要掌握标准函数的级数展开，例如 eˣ、sin x、cos x、ln(1+x) 和 (1+x)ⁿ。理解公式及其应用能有效提高你的考试成绩。

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

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)/2! x² + f”'(0)/3! x³ + … + f⁽ⁿ⁾(0)/n! xⁿ + … . The series uses the values of all derivatives evaluated at zero.

对于在 x = 0 处无限次可导的函数 f(x)，其麦克劳林级数为：f(x) = f(0) + f'(0)x + f”(0)/2! x² + f”'(0)/3! x³ + … + f⁽ⁿ⁾(0)/n! xⁿ + … 。该级数使用了所有导数在零点的取值。

It is a special case of the Taylor series centred at a = 0. The coefficients are determined solely by the behaviour of the function and its derivatives at the origin.

这是泰勒级数在中心 a = 0 处的特例。系数完全由函数及其导数在原点的性质决定。

f(x) = Σₙ₌₀∞ f⁽ⁿ⁾(0) xⁿ / n!

2. Deriving a Maclaurin Expansion Step by Step | 逐步推导麦克劳林展开

To obtain the Maclaurin series of a function, find f(0), f'(0), f”(0), f”'(0), and so on. Substitute these values into the general formula. Look for a pattern in the derivatives to write the general term.

要获得一个函数的麦克劳林级数，先求 f(0)、f'(0)、f”(0)、f”'(0) 等。将这些值代入通式。观察导数的规律以写出通项。

For example, consider f(x) = eˣ. All derivatives are eˣ, so f⁽ⁿ⁾(0) = 1 for all n. The series becomes 1 + x + x²/2! + x³/3! + … .

例如，考虑 f(x) = eˣ。所有导数都是 eˣ，因此对所有 n 都有 f⁽ⁿ⁾(0) = 1。级数变为 1 + x + x²/2! + x³/3! + … 。

3. Standard Expansions You Must Know | 必记的标准展开式

CCEA exams frequently test the expansions of these functions. Memorising them saves time and helps you check your work. The table below lists the essential Maclaurin series for the most common functions.

CCEA 考试经常考查这些函数的展开。熟记它们能节省时间并帮助你检查答案。下表中列出了最常见函数的基本麦克劳林级数。

Function	Maclaurin series (first few terms)
eˣ	1 + x + x²/2! + x³/3! + x⁴/4! + …
sin x	x − x³/3! + x⁵/5! − x⁷/7! + …
cos x	1 − x²/2! + x⁴/4! − x⁶/6! + …
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! + … (\|x\| < 1)

Notice that sin x contains only odd powers, while cos x contains only even powers. The expansion for ln(1+x) does not have factorials in the denominators, which is a common source of mistakes.

注意 sin x 只含奇次幂，cos x 只含偶次幂。ln(1+x) 的展开中分母没有阶乘，这是常见的易错点。

4. Validity and Convergence Intervals | 有效性与收敛区间

Not all Maclaurin series converge for all x. The series for eˣ, sin x and cos x converge for every real x. However, ln(1+x) converges only for −1 < x ≤ 1, and (1+x)ⁿ converges for |x| < 1, unless n is a non‑negative integer which gives a finite binomial expansion valid for all x.

并非所有麦克劳林级数都对一切 x 收敛。eˣ、sin x 和 cos x 的级数对所有实数 x 收敛。而 ln(1+x) 仅在 −1 < x ≤ 1 收敛，(1+x)ⁿ 当 |x| < 1 时收敛，除非 n 为非负整数，此时为有限二项展开，对所有 x 有效。

In CCEA questions you are often asked to state the range of values of x for which the expansion is valid. Always check whether the series is infinite or finite, and apply the appropriate condition.

在 CCEA 考题中，经常要求说明展开有效的 x 的取值范围。务必判断级数是无限还是有限，并应用相应的条件。

5. Finding Coefficients Using Differentiation | 用微分求系数

You may be required to find specific terms without knowing the full series. For instance, to find the coefficient of x³ in the Maclaurin expansion of e^(sin x), you need to compute the third derivative at 0. Differentiate step by step, and evaluate f”'(0).

你可能需要在不了解完整级数的情况下求出特定项。例如，要求 e^(sin x) 麦克劳林展开中 x³ 的系数，需要计算 0 处的三阶导数。逐步微分并求 f”'(0)。

Use the chain rule and product rule carefully. Simplify before evaluating at zero. For composite functions like e^(g(x)), f'(x) = g'(x) e^(g(x)). Then find f”(x) and f”'(x).

仔细运用链式法则和乘法法则。在代入零点前先化简。对于像 e^(g(x)) 这样的复合函数，f'(x) = g'(x) e^(g(x))，再求 f”(x) 和 f”'(x)。

6. Using Substitution to Obtain New Expansions | 用代换法求新展开

You can derive new series by substituting into known expansions. For example, to expand e²ˣ, replace x by 2x in the series for eˣ: 1 + (2x) + (2x)²/2! + (2x)³/3! + … = 1 + 2x + 2x² + (4/3)x³ + … .

你可以通过代入已知展开式来得到新级数。例如，要展开 e²ˣ，将 eˣ 级数中的 x 替换为 2x：1 + (2x) + (2x)²/2! + (2x)³/3! + … = 1 + 2x + 2x² + (4/3)x³ + … 。

Similarly, to expand sin(3x), use the series for sin x and replace x with 3x: 3x − (3x)³/3! + (3x)⁵/5! − … = 3x − (27/6)x³ + … . Always ensure the substitution respects the interval of convergence, but for sin and cos this is automatically true for all real arguments.

类似地，展开 sin(3x) 时，用 sin x 的级数并将 x 替换为 3x：3x − (3x)³/3! + (3x)⁵/5! − … = 3x − (27/6)x³ + … 。务必确保代换后仍满足收敛区间，不过 sin 和 cos 对所有实数自变量都自动成立。

7. Approximations and Error Estimation | 近似与误差估计

In practice, we truncate the series after a few terms to obtain polynomial approximations. For example, using the first three terms of eˣ, e⁰·¹ ≈ 1 + 0.1 + (0.1)²/2 = 1.105. The next term (0.1)³/6 ≈ 0.000167 gives an indication of the error.

实际应用中，我们截取级数的前几项得到多项式近似。例如，用 eˣ 的前三项，e⁰·¹ ≈ 1 + 0.1 + (0.1)²/2 = 1.105。下一项 (0.1)³/6 ≈ 0.000167 可以指示误差大小。

For alternating series such as sin x, the error after truncating is less than the magnitude of the first omitted term. This property is often examined in CCEA questions when estimating accuracy.

对于像 sin x 这样的交错级数，截断后的误差小于第一个被省略项的绝对值。这个性质在 CCEA 考查精度估计时经常出现。

8. Differentiating and Integrating Series Term‑by‑Term | 逐项微分与积分

Within the interval of convergence, Maclaurin series can be differentiated and integrated term‑by‑term. For example, differentiating the series for sin x gives 1 − x²/2! + x⁴/4! − … which is the series for cos x, confirming (sin x)’ = cos x.

在收敛区间内，麦克劳林级数可以逐项微分和积分。例如，对 sin x 的级数逐项微分得到 1 − x²/2! + x⁴/4! − …，这正是 cos x 的级数，验证了 (sin x)’ = cos x。

This technique is useful when you need the series for a function that is the derivative or integral of a known expansion. You can also integrate to find series for functions like arctan x, though that appears more often at A level.

当需要求某个函数的级数，而该函数是已知展开的导数或积分时，这个技巧很有用。你也可以通过积分求 arctan x 等函数的级数，不过这更多出现在 A level 中。

9. Common Exam Question Types | 常见考试题型

(a) ‘Find the Maclaurin series up to the term in x³.’ This requires calculating f(0), f'(0), f”(0) and f”'(0).

(a) “求麦克劳林级数直到 x³ 项。” 这需要计算 f(0)、f'(0)、f”(0) 和 f”'(0)。

(b) ‘Use the expansion to approximate a value.’ Substitute a small value of x into the truncated series and compute the approximate result.

(b) “用展开式近似某个值。” 将 x 的一个较小值代入截断的级数，并计算近似结果。

(c) ‘State the range of validity.’ Quote the interval for which the infinite series converges to the function.

(c) “写出有效范围。” 指出无穷级数收敛到该函数的区间。

(d) ‘Find the coefficient of xⁿ.’ Use pattern recognition or the general term f⁽ⁿ⁾(0)/n!.

(d) “求 xⁿ 的系数。” 利用模式识别或通项 f⁽ⁿ⁾(0)/n!。

10. Spotting and Avoiding Common Mistakes | 识别与避免常见错误

Forgetting to divide by the factorial: f”(0) must be divided by 2! (=2). Many students write f”(0)x² instead of f”(0)x²/2.

忘记除以阶乘：f”(0) 必须除以 2! (=2)。很多学生写成 f”(0)x² 而非 f”(0)x²/2。
Mixing up the expansions for sin x and cos x. Remember sin x starts with x, cos x starts with 1.

混淆 sin x 和 cos x 的展开。记住 sin x 以 x 开头，cos x 以 1 开头。
Applying the binomial series (1+x)ⁿ with n not a positive integer and forgetting the validity condition |x| < 1.

应用二项级数 (1+x)ⁿ 且 n 不是正整数时，忘记有效性条件 |x| < 1。
Incorrectly evaluating derivatives at zero, especially when dealing with products or compositions.

错误计算零点处的导数，尤其是在处理乘积或复合函数时。
Using the substitution method without adjusting the validity interval properly, e.g., stating that the expansion for ln(1+2x) is valid for all x because ln(1+x) is valid for −1 < x ≤ 1. Actually, ln(1+2x) requires −1 < 2x ≤ 1, i.e., −½ < x ≤ ½.

使用代换法时没有正确调整有效性区间，例如说 ln(1+2x) 的展开对一切 x 有效是因为 ln(1+x) 的有效区间是 −1 < x ≤ 1。实际上 ln(1+2x) 要求 −1 < 2x ≤ 1，即 −½ < x ≤ ½。

11. Worked Example: Expand f(x) = e^(2x) sin x up to x³ | 例题：展开 f(x) = e^(2x) sin x 至 x³ 项

Instead of differentiating directly four times, we can multiply the known series. e^(2x) = 1 + 2x + 2x² + (4/3)x³ + … and sin x = x − x³/6 + … . Multiply and collect terms up to x³.

与其直接求四次导数，我们可以将已知级数相乘。e^(2x) = 1 + 2x + 2x² + (4/3)x³ + … 且 sin x = x − x³/6 + … 。相乘并收集不超过 x³ 的项。

(1 + 2x + 2x² + (4/3)x³)(x − x³/6) = 1·x + 1·(−x³/6) + 2x·x + 2x·(−x³/6) + 2x²·x + (4/3)x³·x, ignoring terms beyond x³.

(1 + 2x + 2x² + (4/3)x³)(x − x³/6) = 1·x + 1·(−x³/6) + 2x·x + 2x·(−x³/6) + 2x²·x + (4/3)x³·x，忽略 x³ 以上的项。

This gives x − x³/6 + 2x² − (2/6)x⁴ + 2x³ + … = x + 2x² + (2 − 1/6)x³ + … = x + 2x² + (11/6)x³. Thus f(x) ≈ x + 2x² + (11/6)x³.

得到 x − x³/6 + 2x² − (2/6)x⁴ + 2x³ + … = x + 2x² + (2 − 1/6)x³ + … = x + 2x² + (11/6)x³。因此 f(x) ≈ x + 2x² + (11/6)x³。

12. Final Tips for the Exam | 考试最后的建议

Always show all derivatives up to the required order and evaluate them at 0. Marks are awarded for method.

务必展示所有所需阶数的导数并在 0 处求值。过程分很重要。
If the question states ‘up to the term in x³’, you typically need terms up to f”'(0)x³/3!. Do not stop at x² unless the next term vanishes.

如果问题要求“至 x³ 项”，通常需要写到 f”'(0)x³/3! 项。不要停在 x²，除非下一项为零。
Check validity when the function involves ln or fractional powers. Write the interval in the answer.

当函数包含自然对数或分数次幂时，检查有效性。在答案中写出区间。
Use the Maclaurin series to check limits, approximations and differential equations. Understanding its power will help you connect different topics.

利用麦克劳林级数检查极限、近似值和微分方程。理解其威力有助于贯通不同知识领域。

Published by TutorHao | CCEA GCSE Further Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Maclaurin Series Expansion | 麦克劳林展开考点精讲