📚 GCSE WJEC Maths: Maclaurin Series Key Points | GCSE WJEC 数学：麦克劳林展开 考点精讲
Maclaurin series is a powerful tool for expressing functions as infinite polynomials. For WJEC GCSE Further Mathematics or Level 2 Additional Maths, you are expected to find series expansions up to a few terms and use them for approximations.
麦克劳林级数是将函数表示为无穷多项式的重要工具。对于 WJEC GCSE 进阶数学或附加数学，你需要掌握如何展开到几项，并用于近似计算。
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 about 0. It represents a function f(x) as an infinite sum of terms calculated from the derivatives of f at 0.
麦克劳林级数是函数在 0 处的泰勒级数展开。它将函数 f(x) 表示为从 f 在 0 处的各阶导数计算出的无穷项之和。
ives at x = 0, each divided by the corresponding factorial and multiplied by a power of x.
通用公式利用了函数及其导数在 x = 0 处的值，每一项除以相应的阶乘并乘以 x 的幂。
f(x) = f(0) + f′(0)x + f″(0)x²/2! + f‴(0)x³/3! + … + f⁽ⁿ⁾(0)xⁿ/n! + …
The term f⁽ⁿ⁾(0) means the nth derivative evaluated at 0, and n! denotes the factorial of n. This expansion turns a function into a polynomial of infinite degree.
f⁽ⁿ⁾(0) 表示在 x=0 处的第 n 阶导数，n! 表示 n 的阶乘。该展开将函数转化为无穷次多项式。
2. Standard Maclaurin Series You Must Know | 必须掌握的标准麦克劳林级数
Memorising the basic expansions saves time in WJEC exams. Each one has its own pattern of coefficients and alternating signs.
熟记基本展开能为 WJEC 考试节省时间。每一项都有各自的系数规律和正负号交替模式。
Exponential function: eˣ = 1 + x + x²/2! + x³/3! + x⁴/4! + … for all real x.
指数函数： eˣ = 1 + x + x²/2! + x³/3! + x⁴/4! + …，对所有实数 x 成立。
Sine function: sin x = x − x³/3! + x⁵/5! − x⁷/7! + … for all real x.
正弦函数： sin x = x − x³/3! + x⁵/5! − x⁷/7! + …，对所有实数 x 成立。
Cosine function: cos x = 1 − x²/2! + x⁴/4! − x⁶/6! + … for all real x.
余弦函数： cos x = 1 − x²/2! + x⁴/4! − x⁶/6! + …，对所有实数 x 成立。
Natural logarithm: ln(1 + x) = x − x²/2 + x³/3 − x⁴/4 + … valid for −1 < x ⩽ 1.
自然对数： ln(1 + x) = x − x²/2 + x³/3 − x⁴/4 + …，有效区间为 −1 < x ⩽ 1。
Binomial series: (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.
二项级数： (1 + x)ⁿ = 1 + nx + n(n−1)x²/2! + n(n−1)(n−2)x³/3! + …，当 n 不是正整数时要求 |x| < 1。
3. Step-by-Step Derivation Using Differentiation | 使用微分逐步推导展开
) repeatedly and evaluate each derivative at x = 0. Then plug the values into the formula.
若要从头构建麦克劳林级数，需反复对 f(x) 求导，并在 x=0 处计算各阶导数值，再代入公式。
Worked example: f(x) = e²ˣ
示例：f(x) = e²ˣ
Step 1: f(0) = e⁰ = 1.
Step 2: f′(x) = 2e²ˣ ⇒ f′(0) = 2.
Step 3: f″(x) = 4e²ˣ ⇒ f″(0) = 4.
Step 4: f‴(x) = 8e²ˣ ⇒ f‴(0) = 8.
步骤1: f(0) = e⁰ = 1。
步骤2: f′(x) = 2e²ˣ ⇒ f′(0) = 2。
步骤3: f″(x) = 4e²ˣ ⇒ f″(0) = 4。
步骤4: f‴(x) = 8e²ˣ ⇒ f‴(0) = 8。
Now substitute: f(x) ≈ 1 + 2x + 4x²/2! + 8x³/3! = 1 + 2x + 2x² + 4x³/3 + …
代入得：f(x) ≈ 1 + 2x + 4x²/2! + 8x³/3! = 1 + 2x + 2x² + 4x³/3 + …
WJEC questions often stop at the x³ or x⁴ term. Always simplify the factorial division.
WJEC 考题通常停在 x³ 或 x⁴ 项。一定要化简阶乘除法。
4. Eˣ Expansion and Its Applications | eˣ 展开及其应用
The series for eˣ converges for all real x, making it extremely reliable for approximations. Even a few terms give good accuracy when x is small.
eˣ 的级数对所有实数 x 收敛，因此用于近似非常可靠。当 x 很小时，只用几项就能达到良好精度。
To find e⁰·¹, we use eˣ ≈ 1 + x + x²/2! + x³/3!. With x = 0.1: 1 + 0.1 + 0.01/2 + 0.001/6 = 1.1051666… The exact value is about 1.1051709, giving an error of only 0.0004%.
求 e⁰·¹，可用 eˣ ≈ 1 + x + x²/2! + x³/3!。代入 x = 0.1：1 + 0.1 + 0.01/2 + 0.001/6 = 1.1051666…，精确值约 1.1051709，误差仅 0.0004%。
In WJEC exams you may be asked to calculate such approximations and then find the percentage error.
WJEC 考试中可能要求计算此类近似值，再求出百分误差。
Percentage error = |approximate − exact| / exact × 100%. Always keep exact value from calculator to avoid rounding penalties.
百分误差 = |近似值 − 精确值| / 精确值 × 100%。始终保留计算器精确值以避免四舍五入扣分。
5. Sin x and Cos x Expansions | sin x 与 cos x 的展开
The sine and cosine series alternate in sign and contain only odd or even powers respectively. Their derivatives cycle every four steps, which makes finding higher-order terms systematic.
正弦和余弦级数正负号交替，且分别只含奇次或偶次幂。它们的导数每四步循环一次，这使得求高阶项很有条理。
For sin x: f(0)=0, f′(0)=1, f″(0)=0, f‴(0)=−1, f⁽⁴⁾(0)=0, … giving the series x − x³/3! + x⁵/5! − ….
对 sin x：f(0)=0, f′(0)=1, f″(0)=0, f‴(0)=−1, f⁽⁴⁾(0)=0, … 得到级数 x − x³/3! + x⁵/5! − …。
For cos x: f(0)=1, f′(0)=0, f″(0)=−1, f‴(0)=0, f⁽⁴⁾(0)=1, … giving 1 − x²/2! + x⁴/4! − ….
对 cos x：f(0)=1, f′(0)=0, f″(0)=−1, f‴(0)=0, f⁽⁴⁾(0)=1, … 得到 1 − x²/2! + x⁴/4! − …。
culator. For instance, sin(0.2) ≈ 0.2 − (0.2)³/6 = 0.198667, very close to the true value.
这些展开可用于不用计算器近似三角函数值。例如 sin(0.2) ≈ 0.2 − (0.2)³/6 = 0.198667，非常接近真值。
6. Ln(1+x) Expansion and Interval of Validity | ln(1+x) 展开与有效区间
The Maclaurin series for ln(1+x) is derived from the derivative 1/(1+x), which expands as a geometric series. Integrating term-by-term gives the log series.
ln(1+x) 的麦克劳林级数源于其导数 1/(1+x)，后者可展开为几何级数。逐项积分即得对数级数。
The series
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