IB Mathematics: Analysis & Approaches HL · 鼎睿学苑

Unit E6: Maclaurin
Series单元 E6：麦克劳林级数

The capstone of HL calculus. A Maclaurin series is the polynomial that a smooth function would be if its derivatives at zero were Taylor coefficients. The unit is built on five memorised series ($e^x$, $\sin x$, $\cos x$, $\ln(1+x)$, $(1+x)^n$) and three ways to extend them: substitution, term-by-term differentiation and integration, and combination with L'Hopital's rule for $0/0$ limits.本单元是 HL 微积分的收束。麦克劳林级数把光滑函数展成多项式，零点处的各阶导数即为泰勒系数。建立在五个必背级数（$e^x$、$\sin x$、$\cos x$、$\ln(1+x)$、$(1+x)^n$）之上，再用三条扩展线索：代换、逐项求导与积分，以及与洛必达法则配合处理 $0/0$ 极限。

IB AA HL · Topic AHL 5.18 Papers 1 · 2 · 3 6 Concepts · HL only6 个核心概念 · HL 专属
Read me first先读这里

How to use this guide本指南使用说明

Every section in Unit E6 carries the HL flag. Maclaurin series is an HL-only super-topic; SL students do not see it. The unit reads in two passes: first memorise the five standard series in E6.1, then learn the three derivation techniques (E6.2, E6.3) and the two application techniques (E6.4, E6.6) that build on them. L'Hopital's rule (E6.5) is the classical alternative to E6.6 and is treated in parallel.E6 全单元都带 HL 标记。麦克劳林级数为 HL 专属，SL 不涉及。本单元按两遍阅读：先把 E6.1 的五个标准级数背熟，再掌握三种推导技巧（E6.2、E6.3）和两种应用技巧（E6.4、E6.6）。洛必达法则（E6.5）作为 E6.6 的经典替代，并列处理。

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If you are cramming如果你在临阵磨枪

Memorise the five standard series cold. Practise one substitution problem ($e^{-x^2}$) and one $0/0$ limit problem ($\lim_{x \to 0} (1 - \cos x)/x^2$). Memorise the general Maclaurin formula $f(x) = \sum f^{(n)}(0) x^n / n!$ in case the question asks you to derive a series the standard list does not cover.

把五个标准级数背得滚瓜烂熟。完整做一道代换题（$e^{-x^2}$）和一道 $0/0$ 极限题（$\lim_{x \to 0} (1 - \cos x)/x^2$）。背熟通项公式 $f(x) = \sum f^{(n)}(0) x^n / n!$，以备题目要你推导标准表外的级数。

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If you are going for a 7如果你目标是 7 分

For each standard series, know its radius of convergence and the conditions on $x$ for the equality to hold. For L'Hopital, justify the indeterminate form ($0/0$ or $\infty/\infty$) before differentiating top and bottom. For limits via series, keep enough terms so that the leading non-zero behaviour survives after cancellation.

每个标准级数都要记住收敛半径与等式成立的 $x$ 条件。用洛必达前，必须先论证不定型（$0/0$ 或 $\infty/\infty$）。用级数求极限时，保留的项数要足够，使主导的非零行为在抵消后仍然存在。

HL flagHL 标记说明 All six sections (E6.1 through E6.6) are HL only. The entire super-topic falls under AHL 5.18 in the 2029 IB Subject Brief.六个小节（E6.1 至 E6.6）全部为 HL 专属。整个超级主题在 2029 IB Subject Brief 中归 AHL 5.18。
Topic E6.1 · Standard Maclaurin SeriesTopic E6.1 · 标准麦克劳林级数

Standard Maclaurin Series标准麦克劳林级数 HL AHL 5.18

The five standard series (memorise cold). $$ e^{x} \;=\; \sum_{n=0}^{\infty} \frac{x^{n}}{n!} \;=\; 1 + x + \tfrac{x^{2}}{2!} + \tfrac{x^{3}}{3!} + \cdots, \qquad \text{all real } x. $$ $$ \sin x \;=\; \sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n+1}}{(2n+1)!} \;=\; x - \tfrac{x^{3}}{3!} + \tfrac{x^{5}}{5!} - \cdots, \qquad \text{all real } x. $$ $$ \cos x \;=\; \sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n}}{(2n)!} \;=\; 1 - \tfrac{x^{2}}{2!} + \tfrac{x^{4}}{4!} - \cdots, \qquad \text{all real } x. $$ $$ \ln(1 + x) \;=\; \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{n}}{n} \;=\; x - \tfrac{x^{2}}{2} + \tfrac{x^{3}}{3} - \cdots, \qquad -1 < x \le 1. $$ $$ (1 + x)^{n} \;=\; \sum_{k=0}^{\infty} \binom{n}{k} x^{k} \;=\; 1 + n x + \tfrac{n (n - 1)}{2!} x^{2} + \cdots, \qquad |x| < 1 \text{ if } n \text{ is non-integer}. $$
The general Maclaurin formula. If $f$ has derivatives of all orders at $0$ and the series converges to $f(x)$ on some interval around $0$, then $$ f(x) \;=\; \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^{n} \;=\; f(0) + f'(0) x + \tfrac{f''(0)}{2!} x^{2} + \tfrac{f'''(0)}{3!} x^{3} + \cdots. $$
五个标准级数（必须背熟）。 $$ e^{x} \;=\; \sum_{n=0}^{\infty} \frac{x^{n}}{n!} \;=\; 1 + x + \tfrac{x^{2}}{2!} + \tfrac{x^{3}}{3!} + \cdots, \qquad \text{全体实 } x。 $$ $$ \sin x \;=\; \sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n+1}}{(2n+1)!} \;=\; x - \tfrac{x^{3}}{3!} + \tfrac{x^{5}}{5!} - \cdots, \qquad \text{全体实 } x。 $$ $$ \cos x \;=\; \sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n}}{(2n)!} \;=\; 1 - \tfrac{x^{2}}{2!} + \tfrac{x^{4}}{4!} - \cdots, \qquad \text{全体实 } x。 $$ $$ \ln(1 + x) \;=\; \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{n}}{n} \;=\; x - \tfrac{x^{2}}{2} + \tfrac{x^{3}}{3} - \cdots, \qquad -1 < x \le 1。 $$ $$ (1 + x)^{n} \;=\; \sum_{k=0}^{\infty} \binom{n}{k} x^{k} \;=\; 1 + n x + \tfrac{n (n - 1)}{2!} x^{2} + \cdots, \qquad |x| < 1 \text{（$n$ 非整数时）}。 $$
通项公式（麦克劳林公式）。若 $f$ 在 $0$ 处各阶可导，且级数在 $0$ 的某邻域内收敛于 $f(x)$，则 $$ f(x) \;=\; \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^{n} \;=\; f(0) + f'(0) x + \tfrac{f''(0)}{2!} x^{2} + \tfrac{f'''(0)}{3!} x^{3} + \cdots。 $$
Worked Example E6.1 (derive $e^{x}$ from the formula)E6.1 例题（用通项公式推导 $e^{x}$）

Derive the first four terms of the Maclaurin series for $f(x) = e^{x}$ directly from the general formula.用通项公式直接推导 $f(x) = e^{x}$ 麦克劳林级数的前四项。

Compute derivatives at $0$. Because $\tfrac{d}{dx} e^{x} = e^{x}$, every derivative of $f$ is $e^{x}$, so $f^{(n)}(0) = e^{0} = 1$ for all $n \ge 0$.

计算各阶导数在 $0$ 处的值。由于 $\tfrac{d}{dx} e^{x} = e^{x}$，故 $f$ 的所有阶导数都是 $e^{x}$，从而对所有 $n \ge 0$ 都有 $f^{(n)}(0) = e^{0} = 1$。

Substitute into the Maclaurin formula.

代入麦克劳林公式。

$$ e^{x} \;=\; \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^{n} \;=\; \sum_{n=0}^{\infty} \frac{x^{n}}{n!}. $$

First four terms.

前四项。

$$ e^{x} \;=\; 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \cdots \;=\; 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \cdots. $$

Remark. The series converges for all real $x$ because the ratio of successive terms $|x|/(n+1) \to 0$, regardless of $x$.

注。该级数对全体实 $x$ 收敛，因为相邻两项之比 $|x|/(n+1) \to 0$（与 $x$ 无关）。

Going deeper: why the formula has this shape深入探讨：为何公式呈此形式

Suppose $f(x) = \sum_{n=0}^{\infty} a_{n} x^{n}$ converges in a neighbourhood of $0$. Differentiate term by term $k$ times:

设 $f(x) = \sum_{n=0}^{\infty} a_{n} x^{n}$ 在 $0$ 的邻域内收敛。逐项求导 $k$ 次：

$$ f^{(k)}(x) \;=\; \sum_{n=k}^{\infty} a_{n} \cdot n (n-1) \cdots (n - k + 1) \cdot x^{n - k}. $$

Evaluate at $x = 0$: every term with $n > k$ vanishes, and the term with $n = k$ leaves $a_{k} \cdot k!$. Hence $f^{(k)}(0) = k! \, a_{k}$, which gives $a_{k} = f^{(k)}(0) / k!$. This is the Maclaurin coefficient.

在 $x = 0$ 处求值：$n > k$ 的项全部为零，$n = k$ 的项留下 $a_{k} \cdot k!$。故 $f^{(k)}(0) = k! \, a_{k}$，即 $a_{k} = f^{(k)}(0) / k!$。此即麦克劳林系数。

Caveat. A function can have all derivatives at $0$ yet the resulting series can have radius of convergence zero or can converge to a different function. The standard catalogue of five series is exempt from this pathology: each equals its Maclaurin series on the stated interval.

警告。一个函数虽各阶可导，其麦克劳林级数的收敛半径可能为零，或收敛到别的函数。标准五式不存在此病态：在所述区间上每个函数与其麦克劳林级数恒等。

Pitfall: the radius of convergence陷阱：收敛半径 $e^{x}$, $\sin x$, $\cos x$ converge for all real $x$. The series for $\ln(1 + x)$ is valid only on $-1 < x \le 1$ (the endpoint $x = 1$ converges conditionally; $x = -1$ diverges). The binomial series for $(1 + x)^{n}$ with non-integer $n$ requires $|x| < 1$. State the radius whenever a question asks you to write the series.$e^{x}$、$\sin x$、$\cos x$ 对全体实 $x$ 收敛。$\ln(1 + x)$ 的级数仅在 $-1 < x \le 1$ 上成立（端点 $x = 1$ 条件收敛，$x = -1$ 发散）。$n$ 非整数的 $(1 + x)^{n}$ 二项级数要求 $|x| < 1$。题目要写级数时务必同时写出收敛半径。
The coefficient of $x^{4}$ in the Maclaurin series for $\cos x$ is:$\cos x$ 麦克劳林级数中 $x^{4}$ 的系数为：
E6.1 · Q1
$\dfrac{1}{4!}$
$-\dfrac{1}{4!}$
$\dfrac{1}{24}$
$-\dfrac{1}{24}$
$\cos x = 1 - \tfrac{x^{2}}{2!} + \tfrac{x^{4}}{4!} - \cdots$. The $x^{4}$ coefficient is $\tfrac{1}{4!} = \tfrac{1}{24}$. Options A and C are equal; option C is the simplified numerical form.$\cos x = 1 - \tfrac{x^{2}}{2!} + \tfrac{x^{4}}{4!} - \cdots$。$x^{4}$ 系数为 $\tfrac{1}{4!} = \tfrac{1}{24}$。选项 A 与 C 相等，选项 C 为化简后的数值形式。
In $\cos x = \sum (-1)^{n} x^{2n} / (2n)!$, the $x^{4}$ term corresponds to $n = 2$, giving sign $(-1)^{2} = +1$ and denominator $4! = 24$. Coefficient $= +1/24$.$\cos x = \sum (-1)^{n} x^{2n} / (2n)!$ 中，$x^{4}$ 对应 $n = 2$，符号为 $(-1)^{2} = +1$，分母 $4! = 24$。系数为 $+1/24$。
Topic E6.2 · By SubstitutionTopic E6.2 · 代换法

Series by Substitution into Known Series用代换法导出新级数 HL AHL 5.18

The substitution principle. If $f(u) = \sum_{n=0}^{\infty} a_{n} u^{n}$ is a known series valid for $|u| < R$, then for any well-defined expression $g(x)$ the composed function has Maclaurin series $$ f \bigl( g(x) \bigr) \;=\; \sum_{n=0}^{\infty} a_{n} \bigl[ g(x) \bigr]^{n}, $$ valid wherever $|g(x)| < R$. Substitute $g(x)$ for the dummy variable in the known series and simplify.

Standard substitution targets.
  • $e^{u} \to e^{kx}$, $e^{-x}$, $e^{x^{2}}$, $e^{-x^{2}}$ by $u = kx$, $u = -x$, $u = x^{2}$, $u = -x^{2}$.
  • $\sin u \to \sin (kx)$, $\sin (x^{2})$ similarly.
  • $\dfrac{1}{1 - u} \to \dfrac{1}{1 + x^{2}}$ by $u = -x^{2}$, valid for $|x| < 1$.
The interval of validity transforms with the substitution.
代换原理。设 $f(u) = \sum_{n=0}^{\infty} a_{n} u^{n}$ 是 $|u| < R$ 上的已知级数，则对任意良定的 $g(x)$，复合函数的麦克劳林级数为 $$ f \bigl( g(x) \bigr) \;=\; \sum_{n=0}^{\infty} a_{n} \bigl[ g(x) \bigr]^{n}, $$ 在 $|g(x)| < R$ 处成立。把已知级数里的哑变量换成 $g(x)$ 即可，再化简。

典型代换。
  • $e^{u} \to e^{kx}$、$e^{-x}$、$e^{x^{2}}$、$e^{-x^{2}}$：分别取 $u = kx$、$u = -x$、$u = x^{2}$、$u = -x^{2}$。
  • $\sin u \to \sin (kx)$、$\sin (x^{2})$ 同法。
  • $\dfrac{1}{1 - u} \to \dfrac{1}{1 + x^{2}}$：取 $u = -x^{2}$，在 $|x| < 1$ 上成立。
收敛区间也随代换而变。
Worked Example E6.2 (the Gaussian series for $e^{-x^{2}}$)E6.2 例题（高斯函数 $e^{-x^{2}}$ 的级数）

Find the Maclaurin series for $e^{-x^{2}}$. Write out the first four non-zero terms and the general term.求 $e^{-x^{2}}$ 的麦克劳林级数。写出前四个非零项与通项。

Start from the standard $e^{u}$ series.

由标准 $e^{u}$ 级数出发。

$$ e^{u} \;=\; \sum_{n=0}^{\infty} \frac{u^{n}}{n!}, \qquad \text{all real } u. $$

Substitute $u = -x^{2}$. This is valid for all real $x$ because $-x^{2}$ ranges over all real numbers as $x$ does.

代入 $u = -x^{2}$。当 $x$ 取遍实数时 $-x^{2}$ 亦遍历实数，故对全体实 $x$ 成立。

$$ e^{-x^{2}} \;=\; \sum_{n=0}^{\infty} \frac{(-x^{2})^{n}}{n!} \;=\; \sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n}}{n!}. $$

First four non-zero terms.

前四个非零项。

$$ e^{-x^{2}} \;=\; 1 - x^{2} + \frac{x^{4}}{2!} - \frac{x^{6}}{3!} + \cdots \;=\; 1 - x^{2} + \frac{x^{4}}{2} - \frac{x^{6}}{6} + \cdots. $$

Remark. The function $e^{-x^{2}}$ has no elementary antiderivative; the series form is precisely how the error function $\operatorname{erf}(x) = \tfrac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^{2}} \, dt$ is computed.

注。$e^{-x^{2}}$ 没有初等原函数；误差函数 $\operatorname{erf}(x) = \tfrac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^{2}} \, dt$ 正是通过此级数计算的。

Pitfall: substituting before checking convergence陷阱：未先检查收敛性就代换 If you substitute $u = x$ into $\ln(1 + u) = \sum (-1)^{n+1} u^{n} / n$ to write a series for $\ln(1 + x)$, the original series is valid for $-1 < u \le 1$. Translated, this gives the same interval $-1 < x \le 1$. But if you substitute $u = 2x$, the validity becomes $-1 < 2x \le 1$, that is $-\tfrac{1}{2} < x \le \tfrac{1}{2}$. Always translate the inequality through the substitution.若把 $u = x$ 代入 $\ln(1 + u) = \sum (-1)^{n+1} u^{n} / n$ 得 $\ln(1 + x)$ 的级数，原级数有效区间 $-1 < u \le 1$ 翻译后仍为 $-1 < x \le 1$。但若代 $u = 2x$，有效区间变为 $-1 < 2x \le 1$，即 $-\tfrac{1}{2} < x \le \tfrac{1}{2}$。代换时必须把不等式同时翻译。
The first three non-zero terms of the Maclaurin series for $\sin (2 x)$ are:$\sin (2 x)$ 麦克劳林级数的前三个非零项为：
E6.2 · Q1
$2x - \dfrac{x^{3}}{6} + \dfrac{x^{5}}{120}$
$2 x - \dfrac{4 x^{3}}{3} + \dfrac{4 x^{5}}{15}$
$2 x - \dfrac{8 x^{3}}{6} + \dfrac{32 x^{5}}{120}$
$x - \dfrac{x^{3}}{3} + \dfrac{x^{5}}{15}$
Substitute $u = 2x$ into $\sin u = u - \tfrac{u^{3}}{6} + \tfrac{u^{5}}{120} - \cdots$. Get $\sin (2x) = 2x - \tfrac{(2x)^{3}}{6} + \tfrac{(2x)^{5}}{120} - \cdots = 2 x - \tfrac{8 x^{3}}{6} + \tfrac{32 x^{5}}{120} - \cdots = 2 x - \tfrac{4 x^{3}}{3} + \tfrac{4 x^{5}}{15} - \cdots$. Options B and C are equal; B is the simplified form.把 $u = 2x$ 代入 $\sin u = u - \tfrac{u^{3}}{6} + \tfrac{u^{5}}{120} - \cdots$，得 $\sin (2x) = 2x - \tfrac{(2x)^{3}}{6} + \tfrac{(2x)^{5}}{120} - \cdots = 2 x - \tfrac{8 x^{3}}{6} + \tfrac{32 x^{5}}{120} - \cdots = 2 x - \tfrac{4 x^{3}}{3} + \tfrac{4 x^{5}}{15} - \cdots$。选项 B、C 相等，B 为化简形式。
The substitution $u = 2x$ in $\sin u$ raises each power $u^{k}$ to $(2x)^{k} = 2^{k} x^{k}$. Forgetting to cube and quintuple the $2$ misses the coefficient growth.在 $\sin u$ 中代 $u = 2x$，每个 $u^{k}$ 变成 $(2x)^{k} = 2^{k} x^{k}$。漏掉把 $2$ 立方、五次方就丢了系数。
Topic E6.3 · By Differentiation and IntegrationTopic E6.3 · 求导与积分

Series by Differentiation and Integration用求导与积分导出级数 HL AHL 5.18

The term-by-term theorem. Inside the radius of convergence, a power series can be differentiated and integrated one term at a time: $$ \frac{d}{dx} \sum_{n=0}^{\infty} a_{n} x^{n} \;=\; \sum_{n=1}^{\infty} n \, a_{n} x^{n - 1}, $$ $$ \int \sum_{n=0}^{\infty} a_{n} x^{n} \, dx \;=\; C + \sum_{n=0}^{\infty} \frac{a_{n}}{n + 1} x^{n + 1}. $$ The radius of convergence is preserved (endpoints may change for integration).

The geometric series anchor. $$ \frac{1}{1 - x} \;=\; \sum_{n=0}^{\infty} x^{n} \;=\; 1 + x + x^{2} + x^{3} + \cdots, \qquad |x| < 1. $$ Differentiating or integrating this anchor (and its substitution variants) produces a whole family of series: $\dfrac{1}{(1-x)^{2}}$, $-\ln(1-x)$, $\arctan x$, $\arcsin x$.
逐项定理。在收敛半径内，幂级数可以逐项求导与积分： $$ \frac{d}{dx} \sum_{n=0}^{\infty} a_{n} x^{n} \;=\; \sum_{n=1}^{\infty} n \, a_{n} x^{n - 1}, $$ $$ \int \sum_{n=0}^{\infty} a_{n} x^{n} \, dx \;=\; C + \sum_{n=0}^{\infty} \frac{a_{n}}{n + 1} x^{n + 1}。 $$ 收敛半径保持不变（端点对积分而言可能改变）。

几何级数为根基。 $$ \frac{1}{1 - x} \;=\; \sum_{n=0}^{\infty} x^{n} \;=\; 1 + x + x^{2} + x^{3} + \cdots, \qquad |x| < 1。 $$ 对此根基（及其代换变体）求导或积分可产生一整族级数：$\dfrac{1}{(1-x)^{2}}$、$-\ln(1-x)$、$\arctan x$、$\arcsin x$。
Worked Example E6.3 (integrate the geometric series)E6.3 例题（积分几何级数）

Use the geometric series for $1/(1 - x)$ to derive the Maclaurin series for $-\ln(1 - x)$, and state the interval of validity.用 $1/(1 - x)$ 的几何级数推导 $-\ln(1 - x)$ 的麦克劳林级数，并指出有效区间。

Start with the geometric series.

从几何级数出发。

$$ \frac{1}{1 - x} \;=\; \sum_{n=0}^{\infty} x^{n}, \qquad |x| < 1. $$

Integrate both sides from $0$ to $x$. The antiderivative of $1/(1 - x)$ with respect to $x$ is $-\ln|1 - x|$, and $-\ln|1 - 0| = 0$, so

两边在 $0$ 到 $x$ 上积分。$1/(1 - x)$ 关于 $x$ 的原函数为 $-\ln|1 - x|$，且 $-\ln|1 - 0| = 0$，故

$$ -\ln(1 - x) \;=\; \int_{0}^{x} \frac{dt}{1 - t} \;=\; \int_{0}^{x} \sum_{n=0}^{\infty} t^{n} \, dt \;=\; \sum_{n=0}^{\infty} \frac{x^{n+1}}{n + 1}. $$

Rewrite by shifting the index. Let $m = n + 1$:

换指标改写。令 $m = n + 1$：

$$ -\ln(1 - x) \;=\; \sum_{m=1}^{\infty} \frac{x^{m}}{m} \;=\; x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \frac{x^{4}}{4} + \cdots. $$

Interval. Term-by-term integration preserves the radius of convergence ($|x| < 1$). At the endpoints $x = -1$ and $x = 1$: at $x = -1$, the series becomes $\sum (-1)^{m}/m$, which converges (alternating); at $x = 1$, the series becomes $\sum 1/m$, the harmonic series, which diverges. So the equality holds for $-1 \le x < 1$.

区间。逐项积分保持收敛半径（$|x| < 1$）。端点：$x = -1$ 时级数为 $\sum (-1)^{m}/m$，交错收敛；$x = 1$ 时级数为 $\sum 1/m$（调和级数），发散。故等式在 $-1 \le x < 1$ 上成立。

Consistency check. Replacing $x$ with $-x$ recovers the standard series $\ln(1 + x) = \sum (-1)^{n+1} x^{n} / n$ on $-1 < x \le 1$, exactly as listed in E6.1.

一致性检验。把 $x$ 换为 $-x$ 即可还原 E6.1 列出的标准级数 $\ln(1 + x) = \sum (-1)^{n+1} x^{n} / n$（$-1 < x \le 1$）。

Going deeper: $\arctan x$ from $1/(1 + x^{2})$深入探讨：由 $1/(1 + x^{2})$ 推 $\arctan x$

Substitute $u = -x^{2}$ into the geometric series:

把 $u = -x^{2}$ 代入几何级数：

$$ \frac{1}{1 + x^{2}} \;=\; \frac{1}{1 - (-x^{2})} \;=\; \sum_{n=0}^{\infty} (-x^{2})^{n} \;=\; \sum_{n=0}^{\infty} (-1)^{n} x^{2n}, \qquad |x| < 1. $$

Integrate from $0$ to $x$. Since $\arctan(0) = 0$ and $\tfrac{d}{dx} \arctan x = 1/(1 + x^{2})$:

在 $0$ 到 $x$ 上积分。由 $\arctan(0) = 0$、$\tfrac{d}{dx} \arctan x = 1/(1 + x^{2})$：

$$ \arctan x \;=\; \sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n+1}}{2n + 1} \;=\; x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + \cdots, \qquad |x| \le 1. $$

At $x = 1$, this is the famous Gregory-Leibniz formula $\pi/4 = 1 - 1/3 + 1/5 - \cdots$. The endpoint $x = 1$ is included because the alternating series converges there (Abel's theorem).

$x = 1$ 时即著名的格雷戈里-莱布尼茨公式 $\pi/4 = 1 - 1/3 + 1/5 - \cdots$。端点 $x = 1$ 可纳入，因交错级数在该处收敛（阿贝尔定理）。

Differentiating $\sum_{n=0}^{\infty} x^{n} = 1/(1 - x)$ term by term gives:逐项求导 $\sum_{n=0}^{\infty} x^{n} = 1/(1 - x)$ 得：
E6.3 · Q1
$\dfrac{1}{(1 - x)^{2}} = \sum_{n=1}^{\infty} n x^{n - 1}$
$\dfrac{1}{(1 - x)^{2}} = \sum_{n=0}^{\infty} n x^{n}$
$\dfrac{-1}{(1 - x)^{2}} = \sum_{n=1}^{\infty} n x^{n - 1}$
$\dfrac{1}{1 - x^{2}} = \sum_{n=1}^{\infty} n x^{n - 1}$
By the chain rule, $\tfrac{d}{dx} (1 - x)^{-1} = (1 - x)^{-2}$ (the two minus signs cancel: $-1 \cdot -1 = +1$). Term by term, $\tfrac{d}{dx} x^{n} = n x^{n - 1}$, and the $n = 0$ term ($a_{0} = 1$) differentiates to zero, leaving the sum to start at $n = 1$.由链式法则，$\tfrac{d}{dx} (1 - x)^{-1} = (1 - x)^{-2}$（两个负号抵消：$-1 \cdot -1 = +1$）。逐项 $\tfrac{d}{dx} x^{n} = n x^{n - 1}$；$n = 0$ 项（常数）求导为零，求和从 $n = 1$ 开始。
Differentiating $(1 - x)^{-1}$ gives $(1 - x)^{-2}$ (the minus from the chain rule cancels the exponent's minus). Each $x^{n}$ becomes $n x^{n-1}$, and the constant term differentiates to zero, so the sum starts at $n = 1$.$(1 - x)^{-1}$ 求导得 $(1 - x)^{-2}$（链式负号与指数负号相消）。每个 $x^{n}$ 变成 $n x^{n-1}$，常数项导为零，求和从 $n = 1$ 起。
Topic E6.4 · Approximation and Truncation ErrorTopic E6.4 · 逼近与截断误差

Approximation and Truncation Error逼近与截断误差 HL AHL 5.18

Truncate to approximate. Keeping the first $N + 1$ terms of a Maclaurin series gives the degree-$N$ Maclaurin polynomial: $$ P_{N}(x) \;=\; \sum_{n=0}^{N} \frac{f^{(n)}(0)}{n!} x^{n}. $$ The truncation error $R_{N}(x) = f(x) - P_{N}(x)$ is the sum of all dropped terms.

The dominant-term rule. For $|x|$ small, the error is dominated by the first omitted term, $\tfrac{f^{(N+1)}(0)}{(N+1)!} x^{N+1}$. As $|x|$ grows you need more terms to keep the error small. As $N$ grows (at fixed $x$ inside the radius of convergence) the error tends to zero.

Alternating series bound. If the series is alternating with decreasing absolute terms, the truncation error is at most the absolute value of the first omitted term. This applies cleanly to $\sin x$, $\cos x$, $\ln(1 + x)$ on their convergence intervals.
截断即近似。保留麦克劳林级数前 $N + 1$ 项得到 $N$ 次麦克劳林多项式： $$ P_{N}(x) \;=\; \sum_{n=0}^{N} \frac{f^{(n)}(0)}{n!} x^{n}。 $$ 截断误差 $R_{N}(x) = f(x) - P_{N}(x)$ 是被舍弃项的总和。

主导项原则。$|x|$ 较小时，误差由首个被舍去的项 $\tfrac{f^{(N+1)}(0)}{(N+1)!} x^{N+1}$ 主导。$|x|$ 越大，需保留项数越多。固定 $x$（在收敛半径内）时，$N$ 越大误差越小。

交错级数界。若级数为绝对值递减的交错级数，截断误差不超过首个被舍弃项的绝对值。$\sin x$、$\cos x$、$\ln(1 + x)$ 在其收敛区间上正满足该条件。
Worked Example E6.4 (approximate $e^{0.1}$)E6.4 例题（近似 $e^{0.1}$）

Approximate $e^{0.1}$ using the first four terms of the Maclaurin series for $e^{x}$. Compare with the true value.用 $e^{x}$ 麦克劳林级数的前四项近似 $e^{0.1}$，并与真值比较。

Write the first four terms.

写出前四项。

$$ e^{x} \;\approx\; 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} \;=\; P_{3}(x). $$

Substitute $x = 0.1$.

代入 $x = 0.1$。

$$ P_{3}(0.1) \;=\; 1 + 0.1 + \frac{0.01}{2} + \frac{0.001}{6} \;=\; 1 + 0.1 + 0.005 + 0.000\,166\,6\overline{6}\ldots \;\approx\; 1.105\,167. $$

True value (calculator). $e^{0.1} = 1.105\,170\,918\ldots$. The approximation matches to five decimal places.

真值（计算器）。$e^{0.1} = 1.105\,170\,918\ldots$。近似值与真值前五位小数完全吻合。

Error estimate via the next term. The first omitted term is $\tfrac{x^{4}}{4!} = \tfrac{(0.1)^{4}}{24} = \tfrac{10^{-4}}{24} \approx 4.17 \times 10^{-6}$. This matches the observed error $1.105\,170\,918 - 1.105\,166\,667 \approx 4.25 \times 10^{-6}$ to a fraction of a percent. The dominant-term rule works exactly as advertised.

用下一项估计误差。首个被舍弃项为 $\tfrac{x^{4}}{4!} = \tfrac{(0.1)^{4}}{24} = \tfrac{10^{-4}}{24} \approx 4.17 \times 10^{-6}$。与实际误差 $1.105\,170\,918 - 1.105\,166\,667 \approx 4.25 \times 10^{-6}$ 差不到一个百分点。主导项规则名副其实。

Approximation grows with $|x|$$|x|$ 越大，所需项数越多 The four-term polynomial $P_{3}(x)$ for $e^{x}$ has error $\sim x^{4} / 24$. At $x = 0.1$ this is $4 \times 10^{-6}$, excellent. At $x = 1$ it is $1/24 \approx 0.042$, less than $2\%$. At $x = 2$ it is $16/24 \approx 0.67$, useless (the true value $e^{2} \approx 7.389$ versus $P_{3}(2) = 1 + 2 + 2 + 4/3 \approx 6.333$). Polynomials of fixed degree are local approximations; for large $|x|$ either keep more terms or use a series centred elsewhere.$e^{x}$ 的四项多项式 $P_{3}(x)$ 误差为 $\sim x^{4} / 24$。$x = 0.1$ 时为 $4 \times 10^{-6}$，极佳。$x = 1$ 时为 $1/24 \approx 0.042$，约 $2\%$。$x = 2$ 时为 $16/24 \approx 0.67$，已不可用（真值 $e^{2} \approx 7.389$，$P_{3}(2) = 1 + 2 + 2 + 4/3 \approx 6.333$）。固定次数的多项式是局部近似；$|x|$ 大时要么多保留项，要么改用别处展开的级数。
Using the first three non-zero terms of $\sin x$, the approximate value of $\sin (0.5)$ (to four decimal places) is:用 $\sin x$ 前三个非零项，$\sin (0.5)$ 的近似值（保留四位小数）为：
E6.4 · Q1
$0.5000$
$0.4792$
$0.4794$
$0.5208$
$\sin x \approx x - \tfrac{x^{3}}{6} + \tfrac{x^{5}}{120}$. At $x = 0.5$: $0.5 - \tfrac{0.125}{6} + \tfrac{0.03125}{120} = 0.5 - 0.020833\ldots + 0.000260\ldots = 0.479427\ldots \approx 0.4794$. The true value is $\sin (0.5) = 0.479425\ldots$, agreement to four decimal places.$\sin x \approx x - \tfrac{x^{3}}{6} + \tfrac{x^{5}}{120}$。$x = 0.5$：$0.5 - \tfrac{0.125}{6} + \tfrac{0.03125}{120} = 0.5 - 0.020833\ldots + 0.000260\ldots = 0.479427\ldots \approx 0.4794$。真值 $\sin (0.5) = 0.479425\ldots$，四位小数吻合。
Compute $x - \tfrac{x^{3}}{6} + \tfrac{x^{5}}{120}$ at $x = 0.5$. The cubic term alone gives $0.5 - 0.0208 = 0.4792$; adding the small quintic correction lifts the value to $0.4794$.在 $x = 0.5$ 处算 $x - \tfrac{x^{3}}{6} + \tfrac{x^{5}}{120}$。只到三次项得 $0.5 - 0.0208 = 0.4792$；再加上小的五次修正项升至 $0.4794$。
Topic E6.5 · L'Hopital's RuleTopic E6.5 · 洛必达法则

L'Hopital's Rule洛必达法则 HL AHL 5.18

Statement of the rule. Suppose $f$ and $g$ are differentiable on an open interval around $x = a$ (except possibly at $a$ itself), and $g'(x) \ne 0$ near $a$. If $\lim_{x \to a} f(x) / g(x)$ has the indeterminate form $0/0$ or $\infty / \infty$, then $$ \lim_{x \to a} \frac{f(x)}{g(x)} \;=\; \lim_{x \to a} \frac{f'(x)}{g'(x)}, $$ provided the limit on the right exists (or is $\pm \infty$).

Use the rule when:
  • Direct substitution gives $0/0$ or $\infty / \infty$.
  • The new ratio $f'/g'$ is easier than the original.
Repeat if needed. If $f'(x) / g'(x)$ is still $0/0$ or $\infty / \infty$, apply the rule again. There is no fixed cap on iterations.

Other indeterminate forms. $0 \cdot \infty$ reshape to $0/0$ or $\infty / \infty$ by rewriting; $\infty - \infty$ combines to a single fraction; $1^{\infty}$, $0^{0}$, $\infty^{0}$ take logarithm first.
定理表述。设 $f$ 与 $g$ 在 $x = a$ 的开区间（$a$ 本身可除外）上可导，$g'(x) \ne 0$（在 $a$ 附近）。若 $\lim_{x \to a} f(x) / g(x)$ 为不定型 $0/0$ 或 $\infty / \infty$，则 $$ \lim_{x \to a} \frac{f(x)}{g(x)} \;=\; \lim_{x \to a} \frac{f'(x)}{g'(x)}, $$ 前提是右侧极限存在（或为 $\pm \infty$）。

何时使用：
  • 直接代入得 $0/0$ 或 $\infty / \infty$。
  • 新的比 $f'/g'$ 比原比简单。
必要时反复使用。若 $f'(x) / g'(x)$ 仍为 $0/0$ 或 $\infty / \infty$，再用一次。次数不限。

其他不定型。$0 \cdot \infty$ 改写为 $0/0$ 或 $\infty / \infty$；$\infty - \infty$ 合并为单一分式；$1^{\infty}$、$0^{0}$、$\infty^{0}$ 先取对数。
Worked Example E6.5 (the classic $\sin x / x$ limit)E6.5 例题（经典 $\sin x / x$ 极限）

Evaluate $\lim_{x \to 0} \dfrac{\sin x}{x}$ using L'Hopital's rule.用洛必达法则求 $\lim_{x \to 0} \dfrac{\sin x}{x}$。

Check the indeterminate form. As $x \to 0$, the numerator $\sin x \to 0$ and the denominator $x \to 0$, so the limit has form $0/0$.

检查不定型。$x \to 0$ 时，分子 $\sin x \to 0$、分母 $x \to 0$，故极限为 $0/0$。

Differentiate numerator and denominator separately.

分子分母分别求导。

$$ \lim_{x \to 0} \frac{\sin x}{x} \;=\; \lim_{x \to 0} \frac{(\sin x)'}{(x)'} \;=\; \lim_{x \to 0} \frac{\cos x}{1} \;=\; \cos 0 \;=\; 1. $$

Result. $\lim_{x \to 0} \sin x / x = 1$.

结果。$\lim_{x \to 0} \sin x / x = 1$。

Caveat. Strictly, L'Hopital's rule presupposes the derivative of $\sin x$ is $\cos x$, whose standard proof uses precisely $\lim \sin x / x = 1$. So this is a circular argument as a foundational proof; treat it instead as a quick recheck once the derivative is known. The series approach in E6.6 avoids the circularity.

警告。严格来说，洛必达需要 $\sin x$ 的导数 $\cos x$，而该导数的标准证明恰用 $\lim \sin x / x = 1$。故此论证作为基础证明是循环的；只能视为已知导数后的快速复核。E6.6 的级数法可避免循环。

Going deeper: a repeated-application example深入探讨：反复使用的例子

Evaluate $\lim_{x \to 0} \dfrac{1 - \cos x}{x^{2}}$. As $x \to 0$, both numerator and denominator approach $0$. Apply L'Hopital once:

求 $\lim_{x \to 0} \dfrac{1 - \cos x}{x^{2}}$。$x \to 0$ 时分子分母同趋于 $0$。第一次用洛必达：

$$ \lim_{x \to 0} \frac{1 - \cos x}{x^{2}} \;=\; \lim_{x \to 0} \frac{\sin x}{2 x}. $$

This is still $0/0$. Apply L'Hopital again:

仍为 $0/0$。再用一次：

$$ \lim_{x \to 0} \frac{\sin x}{2 x} \;=\; \lim_{x \to 0} \frac{\cos x}{2} \;=\; \frac{1}{2}. $$

Compare with the series approach in E6.6, which arrives at the same answer in one step.

将此与 E6.6 的级数法比较：后者一步即得同一结果。

Pitfall: applying L'Hopital to a non-indeterminate form陷阱：对非不定型使用洛必达 L'Hopital's rule requires $0/0$ or $\infty/\infty$. Applying it to, say, $\lim_{x \to 0} (\cos x)/(1 + x)$ would give $\lim (-\sin x)/(1) = 0$, which is wrong: direct substitution gives $\cos 0 / (1 + 0) = 1$. Always check the form first. The same trap appears when a problem reduces partway through and the limit ceases to be indeterminate; stop applying the rule the moment substitution gives a finite, defined value.洛必达需要 $0/0$ 或 $\infty/\infty$。若对 $\lim_{x \to 0} (\cos x)/(1 + x)$ 套用，得 $\lim (-\sin x)/(1) = 0$，错。直接代入得 $\cos 0 / (1 + 0) = 1$。必须先检查形式。同陷阱发生在问题做到一半时极限已变为可代入：此时立即停手。
$\lim_{x \to 0} \dfrac{e^{x} - 1}{x}$ equals:$\lim_{x \to 0} \dfrac{e^{x} - 1}{x}$ 等于：
E6.5 · Q1
$0$
$1$
$e$
$\infty$
Form $0/0$. By L'Hopital, $\lim_{x \to 0} (e^{x} - 1)/x = \lim_{x \to 0} e^{x} / 1 = e^{0} = 1$. Equivalent to the derivative of $e^{x}$ at $x = 0$.形式 $0/0$。洛必达：$\lim_{x \to 0} (e^{x} - 1)/x = \lim_{x \to 0} e^{x} / 1 = e^{0} = 1$。等同于 $e^{x}$ 在 $x = 0$ 处的导数。
Check $0/0$, then differentiate top and bottom: $(e^{x} - 1)' = e^{x}$, $(x)' = 1$. The limit is $e^{0} = 1$.先验 $0/0$，再分别求导：$(e^{x} - 1)' = e^{x}$、$(x)' = 1$。极限为 $e^{0} = 1$。
Topic E6.6 · Limits via Maclaurin SeriesTopic E6.6 · 用级数求极限

Limits via Maclaurin Series用麦克劳林级数求极限 HL AHL 5.18

The method. For an indeterminate $0/0$ limit at $x = 0$:
  1. Substitute the Maclaurin series for the numerator and denominator (or for whichever factors are causing the indeterminacy).
  2. Identify the leading non-zero powers of $x$ in numerator and denominator.
  3. Divide numerator and denominator by the lowest common power of $x$ (so the leading term in each becomes a constant).
  4. Take $x \to 0$. Only the constant terms survive.
How many terms to keep. Keep enough so that the leading non-zero term in numerator and denominator both survive after cancellation. For most exam-level problems, three or four terms suffice.

When the series method beats L'Hopital. Series shine when L'Hopital requires multiple applications (e.g. $\lim (1 - \cos x)/x^{2}$ needs two L'Hopital passes; one series substitution gives the answer immediately).
方法。对 $x = 0$ 处的 $0/0$ 极限：
  1. 用麦克劳林级数代换分子分母（或制造不定的部分）。
  2. 找出分子分母中首个非零的 $x$ 次方。
  3. 分子分母同除以最低公共 $x$ 次方，使各自首项化为常数。
  4. 令 $x \to 0$，只剩常数项。
该保留多少项。要保留到分子分母中首个非零项在抵消后仍存在为止。考试题保留三、四项通常够用。

级数法何时胜过洛必达。洛必达需多次重复时级数法显优。例如 $\lim (1 - \cos x)/x^{2}$ 需洛必达两次，而代入级数一次即得。
Worked Example E6.6 ($(1 - \cos x)/x^{2}$ by series)E6.6 例题（用级数求 $(1 - \cos x)/x^{2}$）

Evaluate $\lim_{x \to 0} \dfrac{1 - \cos x}{x^{2}}$ by substituting the Maclaurin series for $\cos x$.用 $\cos x$ 的麦克劳林级数代入求 $\lim_{x \to 0} \dfrac{1 - \cos x}{x^{2}}$。

Substitute the standard series for $\cos x$.

代入 $\cos x$ 的标准级数。

$$ \cos x \;=\; 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + \cdots. $$

Form the numerator.

构造分子。

$$ 1 - \cos x \;=\; 1 - \Bigl[ 1 - \frac{x^{2}}{2} + \frac{x^{4}}{24} - \cdots \Bigr] \;=\; \frac{x^{2}}{2} - \frac{x^{4}}{24} + \frac{x^{6}}{720} - \cdots. $$

Divide by $x^{2}$.

除以 $x^{2}$。

$$ \frac{1 - \cos x}{x^{2}} \;=\; \frac{1}{2} - \frac{x^{2}}{24} + \frac{x^{4}}{720} - \cdots. $$

Take the limit. As $x \to 0$, every term with a positive power of $x$ vanishes:

取极限。$x \to 0$ 时，含正幂的项全部消失：

$$ \lim_{x \to 0} \frac{1 - \cos x}{x^{2}} \;=\; \frac{1}{2}. $$

Comparison. The same answer comes from L'Hopital after two applications (see the "Going deeper" block in E6.5). The series method gets there in one substitution, which is why it dominates in HL exam problems with iterated indeterminacy.

对照。洛必达两次也得同一答案（见 E6.5 的"深入探讨"）。级数法一次代入即可，因此在多次不定型反复出现的 HL 试题中更占优。

Going deeper: how many terms is enough?深入探讨：保留多少项才够？

Suppose you keep only $1 - x^{2}/2$ for $\cos x$. Then $1 - \cos x \approx x^{2}/2$ exactly, and $(1 - \cos x)/x^{2} \to 1/2$. The single dropped term $x^{4}/24$ would, after dividing by $x^{2}$, become $x^{2}/24$, which vanishes as $x \to 0$. So two terms of $\cos x$ are already enough for this problem.

若 $\cos x$ 只保留 $1 - x^{2}/2$，则 $1 - \cos x \approx x^{2}/2$，从而 $(1 - \cos x)/x^{2} \to 1/2$。被舍弃的 $x^{4}/24$ 经除以 $x^{2}$ 后变为 $x^{2}/24$，$x \to 0$ 时为零。故 $\cos x$ 保留两项已够此题。

General principle. Keep enough terms so the leading non-zero behaviour in both numerator and denominator is captured. If a question asks for $\lim (\sin x - x)/x^{3}$, the $\sin x = x - x^{3}/6 + \cdots$ truncated at three terms gives $\sin x - x \approx -x^{3}/6$, hence the limit is $-1/6$. Two terms of $\sin x$ would have given $\sin x - x \approx 0$, which is too coarse.

一般原则。保留到分子分母首个非零行为都能体现为止。若问 $\lim (\sin x - x)/x^{3}$：$\sin x = x - x^{3}/6 + \cdots$ 保留三项给 $\sin x - x \approx -x^{3}/6$，极限为 $-1/6$。只保留两项则 $\sin x - x \approx 0$，太粗。

Using Maclaurin series, $\lim_{x \to 0} \dfrac{\sin x - x}{x^{3}}$ equals:用麦克劳林级数，$\lim_{x \to 0} \dfrac{\sin x - x}{x^{3}}$ 等于：
E6.6 · Q1
$-\dfrac{1}{6}$
$\dfrac{1}{6}$
$0$
$-\dfrac{1}{2}$
$\sin x = x - \tfrac{x^{3}}{6} + \tfrac{x^{5}}{120} - \cdots$, so $\sin x - x = -\tfrac{x^{3}}{6} + \tfrac{x^{5}}{120} - \cdots$. Divide by $x^{3}$: $-\tfrac{1}{6} + \tfrac{x^{2}}{120} - \cdots$. As $x \to 0$, the answer is $-\tfrac{1}{6}$.$\sin x = x - \tfrac{x^{3}}{6} + \tfrac{x^{5}}{120} - \cdots$，故 $\sin x - x = -\tfrac{x^{3}}{6} + \tfrac{x^{5}}{120} - \cdots$。除以 $x^{3}$：$-\tfrac{1}{6} + \tfrac{x^{2}}{120} - \cdots$。$x \to 0$ 时答案 $-\tfrac{1}{6}$。
Substitute $\sin x = x - x^{3}/6 + \cdots$. The leading non-zero contribution to $\sin x - x$ is $-x^{3}/6$. Divide by $x^{3}$ and take the limit: $-1/6$.代 $\sin x = x - x^{3}/6 + \cdots$。$\sin x - x$ 的首个非零贡献为 $-x^{3}/6$。除以 $x^{3}$ 后取极限：$-1/6$。
Exam Tactics考试技巧

Exam Strategy and Common Pitfalls考试策略与常见陷阱

Memorising the five (Paper 1)背熟五式（Paper 1）
  • Write the series, then the validity. Examiners reward both. "$\sin x = x - x^{3}/3! + \cdots$ for all real $x$" earns the full A1, whereas only the formula loses the validity mark.
  • 写级数，也写收敛区间。两者都给分。"$\sin x = x - x^{3}/3! + \cdots$（全体实 $x$）"得满 A1；只写公式漏掉区间要扣分。
  • The signs in $\sin x$ and $\cos x$ alternate; check parity. $\sin$ is odd (only odd powers) and $\cos$ is even (only even powers). A series with both odd and even powers cannot be $\sin x$ or $\cos x$.
  • $\sin x$ 与 $\cos x$ 符号交替；检查奇偶性。$\sin$ 为奇函数（只含奇次幂）、$\cos$ 为偶函数（只含偶次幂）。同时含奇偶次幂的级数不可能是 $\sin x$ 或 $\cos x$。
Substitution and integration (Paper 2)代换与积分（Paper 2）
  • Translate the validity interval through the substitution. $\ln(1 + u)$ valid on $-1 < u \le 1$ becomes $\ln(1 + 2x)$ valid on $-1/2 < x \le 1/2$, not $-1 < x \le 1$.
  • 把收敛区间同步翻译。$\ln(1 + u)$ 在 $-1 < u \le 1$ 有效，故 $\ln(1 + 2x)$ 在 $-1/2 < x \le 1/2$ 有效，不是 $-1 < x \le 1$。
  • Integration introduces a constant of integration. Fix it by evaluating both sides at $x = 0$. For example, integrating $1/(1 - x) = \sum x^{n}$ gives $-\ln(1 - x) = \sum x^{n+1}/(n+1) + C$, and $x = 0$ forces $C = 0$.
  • 积分要引入常数。用 $x = 0$ 两边对照定常数。如积分 $1/(1 - x) = \sum x^{n}$ 得 $-\ln(1 - x) = \sum x^{n+1}/(n+1) + C$，$x = 0$ 给 $C = 0$。
Limits and L'Hopital (Paper 3 HL)极限与洛必达（Paper 3 HL）
  • Verify the indeterminate form first. L'Hopital applies only to $0/0$ or $\infty/\infty$. State the form before differentiating; the M1 is awarded for the verification step.
  • 必须先验不定型。洛必达只用于 $0/0$ 或 $\infty/\infty$。求导前先写出形式；M1 即给在该验证步骤。
  • Pick the right tool. One-step indeterminacies favour L'Hopital; multi-step indeterminacies favour Maclaurin series. For exam answers worth both methods, write the series approach when iterating L'Hopital looks tedious.
  • 选对工具。单次不定型用洛必达；多次反复用级数。当洛必达需多次而显繁琐时，改用级数法作答。
  • Keep enough terms in the series. Underkeeping gives $0/0$ again; overkeeping wastes algebra. The rule is "the lowest power that survives cancellation".
  • 级数保留足够项。少了得 $0/0$，多了浪费代数。规则：保留至抵消后仍存在的最低幂为止。
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$e^{x}$ Maclaurin series?$e^{x}$ 的麦克劳林级数？
$$e^{x} = \sum_{n=0}^{\infty} \frac{x^{n}}{n!}$$All real $x$.全体实 $x$。
$\sin x$ Maclaurin series?$\sin x$ 的麦克劳林级数？
$$\sin x = \sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n+1}}{(2n+1)!}$$All real $x$.全体实 $x$。
$\cos x$ Maclaurin series?$\cos x$ 的麦克劳林级数？
$$\cos x = \sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n}}{(2n)!}$$All real $x$.全体实 $x$。
$\ln(1+x)$ Maclaurin series?$\ln(1+x)$ 的麦克劳林级数？
$$\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{n}}{n}$$$$-1 < x \le 1$$
Binomial $(1+x)^{n}$ series?二项 $(1+x)^{n}$ 级数？
$$(1+x)^{n} = \sum_{k=0}^{\infty} \binom{n}{k} x^{k}$$$|x| < 1$ if $n$ non-integer.$n$ 非整数时 $|x| < 1$。
General Maclaurin formula?麦克劳林通项公式？
$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^{n}$$
Series for $e^{-x^{2}}$?$e^{-x^{2}}$ 的级数？
$$\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n}}{n!}$$Substitute $u = -x^{2}$ in $e^{u}$.在 $e^{u}$ 中代 $u = -x^{2}$。
Geometric series for $1/(1-x)$?$1/(1-x)$ 的几何级数？
$$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^{n}$$$$|x| < 1$$
L'Hopital's rule statement?洛必达法则？
If $0/0$ or $\infty/\infty$:若为 $0/0$ 或 $\infty/\infty$：$$\lim \frac{f}{g} = \lim \frac{f'}{g'}$$
$\lim_{x \to 0} \sin x / x$?$\lim_{x \to 0} \sin x / x$？
$$1$$By L'Hopital or by series: $\sin x = x - x^{3}/6 + \cdots$.用洛必达或级数 $\sin x = x - x^{3}/6 + \cdots$。
$\lim_{x \to 0} (1 - \cos x)/x^{2}$?$\lim_{x \to 0} (1 - \cos x)/x^{2}$？
$$\frac{1}{2}$$From $\cos x = 1 - x^{2}/2 + \cdots$.由 $\cos x = 1 - x^{2}/2 + \cdots$ 得。
Truncation error rule?截断误差规则？
Dominated by the first omitted term: $\sim \dfrac{f^{(N+1)}(0)}{(N+1)!} x^{N+1}$.由首个被舍弃项主导：$\sim \dfrac{f^{(N+1)}(0)}{(N+1)!} x^{N+1}$。
Mixed Practice综合练习

Unit E6 Practice Quiz单元 E6 练习测验

The coefficient of $x^{3}$ in the Maclaurin series for $\sin (2x)$ is:$\sin (2x)$ 麦克劳林级数中 $x^{3}$ 的系数为：
Q1
$-\dfrac{1}{6}$
$-\dfrac{2}{3}$
$-\dfrac{4}{3}$
$\dfrac{4}{3}$
$\sin (2x) = 2x - \tfrac{(2x)^{3}}{3!} + \cdots = 2x - \tfrac{8 x^{3}}{6} + \cdots = 2x - \tfrac{4 x^{3}}{3} + \cdots$. The $x^{3}$ coefficient is $-\tfrac{4}{3}$.$\sin (2x) = 2x - \tfrac{(2x)^{3}}{3!} + \cdots = 2x - \tfrac{8 x^{3}}{6} + \cdots = 2x - \tfrac{4 x^{3}}{3} + \cdots$。$x^{3}$ 系数为 $-\tfrac{4}{3}$。
Sub $u = 2x$ in $\sin u = u - u^{3}/6 + \cdots$: the $u^{3}$ term becomes $(2x)^{3}/6 = 8 x^{3}/6 = 4 x^{3}/3$, with a minus sign.在 $\sin u = u - u^{3}/6 + \cdots$ 中代 $u = 2x$：$u^{3}$ 项变为 $(2x)^{3}/6 = 8 x^{3}/6 = 4 x^{3}/3$，前面有负号。
Using the Maclaurin series for $\ln(1 + x)$, the value of $\ln (1.1)$ to four decimal places is:用 $\ln(1 + x)$ 的麦克劳林级数，$\ln (1.1)$ 保留四位小数为：
Q2
$0.1000$
$0.0953$
$0.1050$
$0.1100$
$\ln(1 + x) = x - x^{2}/2 + x^{3}/3 - \cdots$ At $x = 0.1$: $0.1 - 0.005 + 0.000\overline{3} - 0.000025 + \cdots \approx 0.0953$. True value $\ln (1.1) = 0.0953\,101\ldots$, agreement to four decimal places.$\ln(1 + x) = x - x^{2}/2 + x^{3}/3 - \cdots$。$x = 0.1$：$0.1 - 0.005 + 0.000\overline{3} - 0.000025 + \cdots \approx 0.0953$。真值 $\ln (1.1) = 0.0953\,101\ldots$，四位小数吻合。
Use $\ln(1 + x) = x - x^{2}/2 + x^{3}/3 - x^{4}/4 + \cdots$ at $x = 0.1$. The first four terms give $0.0953$ to four decimal places.用 $\ln(1 + x) = x - x^{2}/2 + x^{3}/3 - x^{4}/4 + \cdots$ 在 $x = 0.1$ 处。前四项得 $0.0953$（四位小数）。
$\lim_{x \to 0} \dfrac{\tan x - x}{x^{3}}$ equals:$\lim_{x \to 0} \dfrac{\tan x - x}{x^{3}}$ 等于：
Q3
$\dfrac{1}{3}$
$\dfrac{1}{6}$
$0$
$-\dfrac{1}{3}$
$\tan x = x + \tfrac{x^{3}}{3} + \tfrac{2 x^{5}}{15} + \cdots$ (this can be derived by dividing $\sin x$ by $\cos x$ as power series). So $\tan x - x = \tfrac{x^{3}}{3} + \tfrac{2 x^{5}}{15} + \cdots$. Divide by $x^{3}$: $\tfrac{1}{3} + \tfrac{2 x^{2}}{15} + \cdots \to \tfrac{1}{3}$. Verifies with L'Hopital (three applications).$\tan x = x + \tfrac{x^{3}}{3} + \tfrac{2 x^{5}}{15} + \cdots$（由 $\sin x$ 除 $\cos x$ 的幂级数得）。故 $\tan x - x = \tfrac{x^{3}}{3} + \tfrac{2 x^{5}}{15} + \cdots$。除以 $x^{3}$：$\tfrac{1}{3} + \tfrac{2 x^{2}}{15} + \cdots \to \tfrac{1}{3}$。洛必达三次亦可验证。
Either expand $\tan x = x + x^{3}/3 + \cdots$ and divide by $x^{3}$, or apply L'Hopital three times. Both give $1/3$.要么展开 $\tan x = x + x^{3}/3 + \cdots$ 后除以 $x^{3}$，要么洛必达三次。均得 $1/3$。
The Maclaurin series for $\dfrac{1}{1 + x^{2}}$ is:$\dfrac{1}{1 + x^{2}}$ 的麦克劳林级数为：
Q4
$1 + x^{2} + x^{4} + x^{6} + \cdots$
$1 - x + x^{2} - x^{3} + \cdots$
$1 - x^{2} + x^{4} - x^{6} + \cdots$
$1 + x + x^{2} + x^{3} + \cdots$
Substitute $u = -x^{2}$ into $1/(1 - u) = \sum u^{n}$: $1/(1 + x^{2}) = \sum (-x^{2})^{n} = \sum (-1)^{n} x^{2n} = 1 - x^{2} + x^{4} - x^{6} + \cdots$, valid for $|x| < 1$. Integrating this gives the $\arctan x$ series.把 $u = -x^{2}$ 代入 $1/(1 - u) = \sum u^{n}$：$1/(1 + x^{2}) = \sum (-x^{2})^{n} = \sum (-1)^{n} x^{2n} = 1 - x^{2} + x^{4} - x^{6} + \cdots$，在 $|x| < 1$ 上有效。对其积分即得 $\arctan x$ 的级数。
Write $1/(1 + x^{2}) = 1/(1 - (-x^{2}))$ and use the geometric series with $u = -x^{2}$. The result alternates in sign and contains only even powers.把 $1/(1 + x^{2})$ 写成 $1/(1 - (-x^{2}))$，再用 $u = -x^{2}$ 的几何级数。结果符号交错且只含偶次幂。
Using L'Hopital's rule, $\lim_{x \to 0} \dfrac{x - \sin x}{x \sin x \cdot x}$ (that is, $\lim (x - \sin x)/(x^{2} \sin x)$) equals:用洛必达，$\lim_{x \to 0} \dfrac{x - \sin x}{x \sin x \cdot x}$（即 $\lim (x - \sin x)/(x^{2} \sin x)$）等于：
Q5
$\dfrac{1}{3}$
$\dfrac{1}{6}$
$0$
$\dfrac{1}{2}$
Replace $\sin x$ in the denominator by its leading term $x$ (valid because $\sin x / x \to 1$): denominator $x^{2} \sin x \sim x^{3}$. The numerator $x - \sin x = x - (x - x^{3}/6 + \cdots) = x^{3}/6 - \cdots$. Hence the limit is $(x^{3}/6) / x^{3} = 1/6$. Confirms via three applications of L'Hopital.把分母中的 $\sin x$ 替为首项 $x$（由 $\sin x / x \to 1$ 合法）：分母 $x^{2} \sin x \sim x^{3}$。分子 $x - \sin x = x - (x - x^{3}/6 + \cdots) = x^{3}/6 - \cdots$。极限为 $(x^{3}/6) / x^{3} = 1/6$。洛必达三次亦得同结果。
Use the series $\sin x = x - x^{3}/6 + \cdots$. Numerator $\sim x^{3}/6$, denominator $\sim x^{3}$. Ratio $\to 1/6$. L'Hopital reaches the same answer after three applications.用级数 $\sin x = x - x^{3}/6 + \cdots$。分子 $\sim x^{3}/6$、分母 $\sim x^{3}$。比 $\to 1/6$。洛必达三次同果。
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IB Math Analysis & Approaches HL · Unit E6: Maclaurin Series · First Assessment 2029IB 数学分析与方法 HL · 单元 E6：麦克劳林级数 · 首次考核 2029