IB Math AA HL · Unit E6: Maclaurin Series · Solutions

Syllabus AHL 5.18 · Maclaurin series, substitution, term-by-term differentiation/integration, L'Hopital, limits via series考纲 AHL 5.18 · 麦克劳林级数、代换、逐项求导/积分、洛必达、用级数求极限AA HL

Section A · Worked SolutionsA 节 · 详细解析

Q1EASYPaper 1AAHL 5.18 Standard Series Recall (HL)[4 marks]

First three terms of the Maclaurin series for $e^{x}$; use them to approximate $e^{0.1}$.写出 $e^{x}$ 麦克劳林级数前三项；用其近似 $e^{0.1}$。

Answers:答案： (a) $1 + x + \tfrac{x^{2}}{2}$ · (b) $e^{0.1} \approx 1.105$

(a) First three terms A1·A1

The standard series is $e^{x} = \sum_{n=0}^{\infty} \dfrac{x^{n}}{n!} = 1 + x + \dfrac{x^{2}}{2!} + \dfrac{x^{3}}{3!} + \cdots$. Constant, linear, quadratic terms: $$ e^{x} \;\approx\; 1 \;+\; x \;+\; \tfrac{x^{2}}{2}. $$

(b) Substitute $x = 0.1$ M1·A1

$$ e^{0.1} \;\approx\; 1 + 0.1 + \tfrac{(0.1)^{2}}{2} \;=\; 1 + 0.1 + 0.005 \;=\; 1.105. $$ As an exact fraction: $1 + \tfrac{1}{10} + \tfrac{1}{200} = \tfrac{200 + 20 + 1}{200} = \tfrac{221}{200}$.

Memorise the five standard series — they unlock everything else. The five HL series ($e^{x}$, $\sin x$, $\cos x$, $\ln(1+x)$, $(1+x)^{k}$) are the keys: every other Maclaurin series in this unit (and on the IB exam) is built by substituting, differentiating, integrating, or multiplying these. Lose them and you fall back to the derivative-table formula — slow and error-prone. Drill them in this order: $e^{x}$ (all $n!$), $\sin x$ (odd powers, alternating), $\cos x$ (even powers, alternating), $\ln(1+x)$ (no factorials, alternating, denominators $1, 2, 3, \ldots$), $(1+x)^{k}$ (binomial coefficients, terminates iff $k \in \mathbb{N}$). True $e^{0.1} = 1.10517\ldots$; our $3$-term estimate is already accurate to $4$ decimal places because the next term $\tfrac{(0.1)^{3}}{6} \approx 0.000167$ is small.

(a) 前三项 A1·A1

标准级数：$e^{x} = \sum_{n=0}^{\infty} \dfrac{x^{n}}{n!} = 1 + x + \dfrac{x^{2}}{2!} + \dfrac{x^{3}}{3!} + \cdots$。常数、一次、二次项： $$ e^{x} \;\approx\; 1 \;+\; x \;+\; \tfrac{x^{2}}{2}. $$

(b) 代入 $x = 0.1$ M1·A1

$$ e^{0.1} \;\approx\; 1 + 0.1 + \tfrac{(0.1)^{2}}{2} \;=\; 1 + 0.1 + 0.005 \;=\; 1.105. $$ 精确分数：$1 + \tfrac{1}{10} + \tfrac{1}{200} = \tfrac{221}{200}$。

五个标准级数必背——是其他一切的钥匙。HL 五个标准级数（$e^{x}$、$\sin x$、$\cos x$、$\ln(1+x)$、$(1+x)^{k}$）是钥匙：本单元（及 IB 考试）所有其他麦克劳林级数都由代换、求导、积分、相乘这五个得到。丢掉它们就只能回到导数表通项公式——慢且易错。按此顺序记：$e^{x}$（所有 $n!$）、$\sin x$（奇次幂，交错）、$\cos x$（偶次幂，交错）、$\ln(1+x)$（无阶乘，交错，分母 $1, 2, 3, \ldots$）、$(1+x)^{k}$（二项式系数，$k \in \mathbb{N}$ 时终止）。真值 $e^{0.1} = 1.10517\ldots$；三项估计已达 $4$ 位小数精度，因为下一项 $\tfrac{(0.1)^{3}}{6} \approx 0.000167$ 很小。

Q2MEDIUMPaper 1AAHL 5.18 Series by Substitution (HL)[5 marks]

$g(x) = e^{-x^{2}}$. Find first 3 non-zero terms by substituting $u = -x^{2}$ into $e^{u}$; give the general term; state convergence.$g(x) = e^{-x^{2}}$。代 $u = -x^{2}$ 入 $e^{u}$ 求前 3 个非零项；给出通项；写出收敛范围。

Answers:答案： (a) $1 - x^{2} + \tfrac{x^{4}}{2}$ · (b) $\dfrac{(-1)^{n} x^{2n}}{n!}$ · (c) all $x \in \mathbb{R}$$x \in \mathbb{R}$

(a) Substitute $u = -x^{2}$ M1·A1·A1

Start from $e^{u} = 1 + u + \tfrac{u^{2}}{2!} + \tfrac{u^{3}}{3!} + \cdots$. Substitute $u = -x^{2}$: $$ e^{-x^{2}} \;=\; 1 + (-x^{2}) + \tfrac{(-x^{2})^{2}}{2!} + \tfrac{(-x^{2})^{3}}{3!} + \cdots \;=\; 1 - x^{2} + \tfrac{x^{4}}{2} - \tfrac{x^{6}}{6} + \cdots. $$ First three non-zero terms: $1 - x^{2} + \tfrac{x^{4}}{2}$.

(b) General term A1

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

(c) Interval of convergence A1

The series for $e^{u}$ converges for every $u \in \mathbb{R}$ (radius of convergence $\infty$). The substitution $u = -x^{2}$ is real for every $x \in \mathbb{R}$, so the substituted series converges to $e^{-x^{2}}$ for every $x \in \mathbb{R}$.

Substitution is the workhorse — and it preserves convergence. The cleanest way to write a Maclaurin series is almost never the derivative-table formula; it is "spot a known $u$-substitution and plug in." Three rules: (i) only even powers appear when $u$ is an even function of $x$ (here $u = -x^{2}$, so the series has only $x^{0}, x^{2}, x^{4}, \ldots$); (ii) signs alternate when the substituted $u$ has a sign — track $(-1)^{n}$ separately from $u^{n}$ to avoid arithmetic slips; (iii) convergence transfers: if the original series converges for $|u| < R$, the substituted series converges for those $x$ with $|u(x)| < R$. For $e^{u}$ with $R = \infty$, the substituted series converges everywhere $u$ is defined. This is exactly why the Gaussian $e^{-x^{2}}$ — which has no elementary antiderivative — is still tractable: its Maclaurin series exists, converges on all of $\mathbb{R}$, and can be integrated term-by-term to define the error function $\operatorname{erf}(x)$.

(a) 代 $u = -x^{2}$ M1·A1·A1

由 $e^{u} = 1 + u + \tfrac{u^{2}}{2!} + \tfrac{u^{3}}{3!} + \cdots$ 出发，代 $u = -x^{2}$： $$ e^{-x^{2}} \;=\; 1 + (-x^{2}) + \tfrac{(-x^{2})^{2}}{2!} + \tfrac{(-x^{2})^{3}}{3!} + \cdots \;=\; 1 - x^{2} + \tfrac{x^{4}}{2} - \tfrac{x^{6}}{6} + \cdots. $$ 前三个非零项：$1 - x^{2} + \tfrac{x^{4}}{2}$。

(b) 通项 A1

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

(c) 收敛区间 A1

$e^{u}$ 级数对任意 $u \in \mathbb{R}$ 收敛（收敛半径 $\infty$）。$u = -x^{2}$ 对任意 $x \in \mathbb{R}$ 为实数，故代换后级数对任意 $x \in \mathbb{R}$收敛于 $e^{-x^{2}}$。

代换是主力——且保收敛。写出麦克劳林级数最干净的方法几乎从不是导数表通项，而是"识别可用的 $u$ 代换并代入"。三条规则：(i) 当 $u$ 是 $x$ 的偶函数时，级数只含偶次幂（本题 $u = -x^{2}$，故级数只有 $x^{0}, x^{2}, x^{4}, \ldots$）；(ii) 当代换的 $u$ 带符号时，要把 $(-1)^{n}$ 和 $u^{n}$ 分开追踪，避免算术失误；(iii) 收敛性平移：若原级数在 $|u| < R$ 收敛，则代换后级数对满足 $|u(x)| < R$ 的 $x$ 收敛。$e^{u}$ 的 $R = \infty$，故代换后处处收敛。这正是为何高斯函数 $e^{-x^{2}}$——无初等原函数——仍可处理：其麦克劳林级数处处收敛，可逐项积分定义误差函数 $\operatorname{erf}(x)$。

Q3MEDIUMPaper 1AAHL 5.18 Term-by-Term Integration (HL)[6 marks]

Series for $\dfrac{1}{1 + x^{2}}$ (first 4 non-zero terms, convergence); integrate term-by-term to get series for $\arctan x$.求 $\dfrac{1}{1 + x^{2}}$ 级数前 4 个非零项与收敛范围；逐项积分得 $\arctan x$ 级数。

Answers:答案： (a) $1 - x^{2} + x^{4} - x^{6}$, $|x| < 1$ · (b) $x - \tfrac{x^{3}}{3} + \tfrac{x^{5}}{5} - \tfrac{x^{7}}{7}$

(a) Geometric series with ratio $-x^{2}$ M1·A1·A1

Write $\dfrac{1}{1 + x^{2}} = \dfrac{1}{1 - (-x^{2})}$. The geometric series $\dfrac{1}{1 - r} = 1 + r + r^{2} + r^{3} + \cdots$ with $r = -x^{2}$: $$ \frac{1}{1 + x^{2}} \;=\; 1 + (-x^{2}) + (-x^{2})^{2} + (-x^{2})^{3} + \cdots \;=\; 1 - x^{2} + x^{4} - x^{6} + \cdots. $$ Convergence: $|r| < 1 \Leftrightarrow |{-x^{2}}| < 1 \Leftrightarrow x^{2} < 1 \Leftrightarrow |x| < 1$.

(b) Integrate term-by-term M1·A1·A1

$\displaystyle\int \dfrac{1}{1 + x^{2}}\, dx = \arctan x + C$. Integrate the series term-by-term (valid inside the radius of convergence): $$ \arctan x + C \;=\; \int \bigl(1 - x^{2} + x^{4} - x^{6} + \cdots\bigr) dx \;=\; x - \tfrac{x^{3}}{3} + \tfrac{x^{5}}{5} - \tfrac{x^{7}}{7} + \cdots + C. $$ Why $C = 0$: evaluate both sides at $x = 0$. LHS: $\arctan 0 + C = 0 + C = C$. RHS: $0 - 0 + 0 - 0 = 0$. Hence $C = 0$ and $$ \arctan x \;=\; x - \tfrac{x^{3}}{3} + \tfrac{x^{5}}{5} - \tfrac{x^{7}}{7} + \cdots. $$

Term-by-term integration is a magic wand — within the radius. Inside the open interval of convergence, a power series can be differentiated or integrated term-by-term, and the new series has the same radius. That is how the Maclaurin series of $\arctan x$ ($|x| < 1$, with conditional convergence at the endpoints), $\ln(1+x)$ (from $\tfrac{1}{1+x}$), and many others are derived without ever computing a single derivative. The "$C = 0$" check at $x = 0$ is non-negotiable: integration introduces an unknown constant, and the only way to fix it is to evaluate both sides at $x = 0$ (the centre of the Maclaurin expansion). For most "nice" functions like $\arctan x$ this gives $C = 0$, but for, e.g., $\int \sec^{2} x\, dx = \tan x + C$, the same trick still applies — $\tan 0 = 0$, so the constant from integrating $1 + x^{2} + \tfrac{2}{3} x^{4} + \cdots$ (the series for $\sec^{2} x$) is also $0$.

(a) 公比为 $-x^{2}$ 的几何级数 M1·A1·A1

把 $\dfrac{1}{1 + x^{2}} = \dfrac{1}{1 - (-x^{2})}$。几何级数 $\dfrac{1}{1 - r} = 1 + r + r^{2} + r^{3} + \cdots$，代 $r = -x^{2}$： $$ \frac{1}{1 + x^{2}} \;=\; 1 - x^{2} + x^{4} - x^{6} + \cdots. $$ 收敛：$|r| < 1 \Leftrightarrow x^{2} < 1 \Leftrightarrow |x| < 1$。

(b) 逐项积分 M1·A1·A1

$\displaystyle\int \dfrac{1}{1 + x^{2}}\, dx = \arctan x + C$。在收敛半径内逐项积分： $$ \arctan x + C \;=\; x - \tfrac{x^{3}}{3} + \tfrac{x^{5}}{5} - \tfrac{x^{7}}{7} + \cdots + C. $$ $C = 0$ 的理由：令 $x = 0$。左边 $= \arctan 0 + C = C$；右边 $= 0$。故 $C = 0$， $$ \arctan x \;=\; x - \tfrac{x^{3}}{3} + \tfrac{x^{5}}{5} - \tfrac{x^{7}}{7} + \cdots. $$

逐项积分是魔法——但限在半径之内。在收敛开区间内，幂级数可逐项求导或积分，新级数收敛半径相同。$\arctan x$（$|x| < 1$，端点条件收敛）、$\ln(1+x)$（由 $\tfrac{1}{1+x}$ 得）等许多级数都用此法导出，无需求任何导数。"$C = 0$" 的核对必不可少：积分会引入未知常数，唯一定它的方式就是把两边代入 $x = 0$（麦克劳林展开的中心）。对 $\arctan x$ 这类"好"函数得 $C = 0$；对 $\int \sec^{2} x\, dx = \tan x + C$ 等同理——$\tan 0 = 0$，故由 $\sec^{2} x$ 级数 $1 + x^{2} + \tfrac{2}{3} x^{4} + \cdots$ 积出的常数也是 $0$。

Q4HARDPaper 1AAHL 5.18 L'Hopital (Twice) (HL)[5 marks]

Evaluate $\displaystyle\lim_{x \to 0} \dfrac{e^{x} - 1 - x}{x^{2}}$ via L'Hopital (twice).用洛必达（两次）求 $\displaystyle\lim_{x \to 0} \dfrac{e^{x} - 1 - x}{x^{2}}$。

Answer:答案： $\dfrac{1}{2}$

(a) Direct substitution: $0/0$ A1

Numerator at $x = 0$: $e^{0} - 1 - 0 = 1 - 1 - 0 = 0$. Denominator at $x = 0$: $0^{2} = 0$. The form is $\boldsymbol{0/0}$, indeterminate.

(b) First L'Hopital M1·A1

Apply L'Hopital's rule: differentiate numerator and denominator separately. $$ \lim_{x \to 0} \frac{e^{x} - 1 - x}{x^{2}} \;=\; \lim_{x \to 0} \frac{e^{x} - 1}{2x}. $$ Check the new limit at $x = 0$: numerator $= e^{0} - 1 = 0$, denominator $= 0$. Still $\boldsymbol{0/0}$.

(c) Second L'Hopital M1·A1

Apply L'Hopital again: $$ \lim_{x \to 0} \frac{e^{x} - 1}{2x} \;=\; \lim_{x \to 0} \frac{e^{x}}{2} \;=\; \frac{e^{0}}{2} \;=\; \frac{1}{2}. $$ Hence the original limit is $\boxed{1/2}$.

Declare the form before every L'Hopital application. The IB markscheme allocates a method mark for "states $0/0$ form" each time you invoke L'Hopital. Skip the declaration and the M1 is gone, even if the algebra is right. Equally important: verify the new ratio is indeterminate before applying L'Hopital again — applying it to a determinate ratio gives a wrong answer. Sanity check by series: $e^{x} - 1 - x = \tfrac{x^{2}}{2} + \tfrac{x^{3}}{6} + \tfrac{x^{4}}{24} + \cdots$, so $\dfrac{e^{x} - 1 - x}{x^{2}} = \tfrac{1}{2} + \tfrac{x}{6} + \tfrac{x^{2}}{24} + \cdots \to \tfrac{1}{2}$ as $x \to 0$. The series method bypasses both L'Hopital applications in one step — this is exactly the "limits via Maclaurin" trick of Q7.

(a) 直接代入：$0/0$ A1

$x = 0$ 时分子 $e^{0} - 1 - 0 = 0$；分母 $0^{2} = 0$。不定型 $\boldsymbol{0/0}$。

(b) 第一次洛必达 M1·A1

对分子分母分别求导： $$ \lim_{x \to 0} \frac{e^{x} - 1 - x}{x^{2}} \;=\; \lim_{x \to 0} \frac{e^{x} - 1}{2x}. $$ 新极限在 $x = 0$：分子 $e^{0} - 1 = 0$，分母 $0$。仍为 $\boldsymbol{0/0}$。

(c) 第二次洛必达 M1·A1

再次洛必达： $$ \lim_{x \to 0} \frac{e^{x} - 1}{2x} \;=\; \lim_{x \to 0} \frac{e^{x}}{2} \;=\; \frac{1}{2}. $$ 原极限 $= \boxed{1/2}$。

每次用洛必达前都要先声明不定型。IB 评分标准为每一次"声明 $0/0$ 形式"给一个方法分（M1）。略过声明就丢 M1，即便后续代数正确。同样关键：再次使用洛必达前必验证新比仍为不定型——对确定比用洛必达会得错答。用级数核对：$e^{x} - 1 - x = \tfrac{x^{2}}{2} + \tfrac{x^{3}}{6} + \cdots$，故 $\dfrac{e^{x} - 1 - x}{x^{2}} = \tfrac{1}{2} + \tfrac{x}{6} + \cdots \to \tfrac{1}{2}$。级数方法一步绕开两次洛必达——这正是 Q7 的"用麦克劳林求极限"。

Section B · Worked SolutionsB 节 · 详细解析

Q5HARDPaper 1BAHL 5.18 Build Series from General Formula (HL)[11 marks]

$f(x) = \ln(1 + 2x)$. Derivative table ($n = 0, 1, 2, 3$); first 4 terms by Maclaurin formula; same by substitution into $\ln(1+u)$; general term.$f(x) = \ln(1 + 2x)$。导数表（$n = 0, 1, 2, 3$）；用麦克劳林公式给出前 4 项；代 $u = 2x$ 入 $\ln(1+u)$ 复核；通项。

Answers:答案： (b) $2x - 2x^{2} + \tfrac{8 x^{3}}{3} - 4 x^{4}$ · (c) $|x| < \tfrac{1}{2}$ · (d) $\dfrac{(-1)^{n+1} (2x)^{n}}{n}$

(a) Derivative table M1·A1·A1·A1·A1

Differentiate $f(x) = \ln(1 + 2x)$ using the chain rule each time.

$n = 0$: $f(x) = \ln(1 + 2x)$, so $f(0) = \ln 1 = 0$. Coefficient $f(0)/0! = 0$.
$n = 1$: $f'(x) = \dfrac{2}{1 + 2x}$, so $f'(0) = 2$. Coefficient $f'(0)/1! = 2$.
$n = 2$: $f''(x) = -\dfrac{4}{(1 + 2x)^{2}}$, so $f''(0) = -4$. Coefficient $f''(0)/2! = -2$.
$n = 3$: $f'''(x) = \dfrac{16}{(1 + 2x)^{3}}$, so $f'''(0) = 16$. Coefficient $f'''(0)/3! = 16/6 = 8/3$.

(b) First four terms via Maclaurin formula M1·A1

$$ f(x) \;=\; \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}\, x^{n} \;=\; 0 + 2x - 2 x^{2} + \tfrac{8}{3}\, x^{3} - 4 x^{4} + \cdots. $$ (The $n = 4$ term: $f^{(4)}(x) = -\dfrac{96}{(1+2x)^{4}}$, $f^{(4)}(0) = -96$, coefficient $-96/24 = -4$.)

(c) Verify by substitution $u = 2x$ M1·A1·A1

Standard series: $\ln(1 + u) = u - \tfrac{u^{2}}{2} + \tfrac{u^{3}}{3} - \tfrac{u^{4}}{4} + \cdots$, valid for $-1 < u \le 1$. Substitute $u = 2x$: $$ \ln(1 + 2x) \;=\; 2x - \tfrac{(2x)^{2}}{2} + \tfrac{(2x)^{3}}{3} - \tfrac{(2x)^{4}}{4} + \cdots \;=\; 2x - 2 x^{2} + \tfrac{8}{3}\, x^{3} - 4 x^{4} + \cdots. \;\checkmark $$ Matches (b). Convergence: need $-1 < 2x \le 1$, i.e. $-\tfrac{1}{2} < x \le \tfrac{1}{2}$. Inside the open interval $|x| < \tfrac{1}{2}$ the convergence is absolute; at $x = \tfrac{1}{2}$ it is conditional.

(d) General term A1

From the standard series, the $n$-th term ($n \ge 1$) is $\dfrac{(-1)^{n+1}\, u^{n}}{n} = \dfrac{(-1)^{n+1}\, (2x)^{n}}{n} = \dfrac{(-1)^{n+1}\, 2^{n}\, x^{n}}{n}$.

Substitution beats the derivative table every time — but the table earns method marks. Compare the two routes. Derivative-table route: $4$ derivatives, $4$ evaluations at $0$, $4$ divisions by $n!$ — about $12$ arithmetic steps with $4$ chances to make a chain-rule sign error. Substitution route: write $\ln(1+u)$ from memory ($1$ step), substitute $u = 2x$ (gives $(2x)^{n}$), simplify each $(2x)^{n}/n$ ($1$ step per term). The substitution route is $\sim 4\times$ faster and far more reliable for any series built from a standard one. So why does the IB include derivative-table questions? Because some functions are not built from standard series (e.g. $\arctan(\sin x)$, or any "made-up" function). You need the derivative table as a fallback. Strategy: identify the route in the first $30$ seconds — if a substitution exists, use it; if not, build the table. And note that the derivative-table marks (M1 for the formula, A1 per derivative) are awarded even if the substitution route would have been cleaner.

(a) 导数表 M1·A1·A1·A1·A1

用链式法则逐次对 $f(x) = \ln(1 + 2x)$ 求导。

$n = 0$：$f(x) = \ln(1 + 2x)$，$f(0) = \ln 1 = 0$。系数 $f(0)/0! = 0$。
$n = 1$：$f'(x) = \dfrac{2}{1 + 2x}$，$f'(0) = 2$。系数 $2$。
$n = 2$：$f''(x) = -\dfrac{4}{(1 + 2x)^{2}}$，$f''(0) = -4$。系数 $-4/2 = -2$。
$n = 3$：$f'''(x) = \dfrac{16}{(1 + 2x)^{3}}$，$f'''(0) = 16$。系数 $16/6 = 8/3$。

(b) 用麦克劳林公式给出前四项 M1·A1

$$ f(x) \;=\; \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}\, x^{n} \;=\; 2x - 2 x^{2} + \tfrac{8}{3}\, x^{3} - 4 x^{4} + \cdots. $$ （第 $4$ 项：$f^{(4)}(x) = -\dfrac{96}{(1+2x)^{4}}$，$f^{(4)}(0) = -96$，系数 $-96/24 = -4$。）

(c) 代换 $u = 2x$ 验证 M1·A1·A1

标准级数 $\ln(1 + u) = u - \tfrac{u^{2}}{2} + \tfrac{u^{3}}{3} - \tfrac{u^{4}}{4} + \cdots$（$-1 < u \le 1$）。代 $u = 2x$： $$ \ln(1 + 2x) \;=\; 2x - 2 x^{2} + \tfrac{8}{3}\, x^{3} - 4 x^{4} + \cdots. \;\checkmark $$ 与 (b) 一致。收敛：需 $-1 < 2x \le 1$，即 $-\tfrac{1}{2} < x \le \tfrac{1}{2}$；开区间 $|x| < \tfrac{1}{2}$ 内绝对收敛，$x = \tfrac{1}{2}$ 处条件收敛。

(d) 通项 A1

由标准级数，第 $n$ 项（$n \ge 1$）为 $\dfrac{(-1)^{n+1}\, (2x)^{n}}{n} = \dfrac{(-1)^{n+1}\, 2^{n}\, x^{n}}{n}$。

代换永远比导数表快——但导数表能拿方法分。对比两条路。导数表路线：$4$ 次求导、$4$ 次代 $0$、$4$ 次除以 $n!$——约 $12$ 步运算，$4$ 次链式法则有可能出符号错。代换路线：默写 $\ln(1+u)$（$1$ 步）、代 $u = 2x$（得 $(2x)^{n}$）、简化 $(2x)^{n}/n$（每项 $1$ 步）。代换比导数表快约 $4$ 倍，对由标准级数衍生的函数远更可靠。那 IB 为何还出导数表题？因为有些函数不是由标准级数衍生（如 $\arctan(\sin x)$，或任意"人造"函数），导数表是兜底。策略：开题 $30$ 秒内决定路线——能代换就代换，不能就建表。注意：即便代换更优，按导数表写出公式（M1）和每个导数（A1）的分仍会给。

Paper 2 · Worked Solutions第二卷 · 详细解析

Q6MEDIUMPaper 2AHL 5.18 Truncated Approximation (HL)[7 marks]

Approximate $\cos(0.5)$ with first 4 terms of $\cos x$ series; compute and compare with GDC.用 $\cos x$ 级数前 4 项近似 $\cos(0.5)$；计算并与 GDC 对比。

Answers:答案： (c) $\cos(0.5) \approx 0.8776$ · (d) absolute error绝对误差 $\approx 9.7 \times 10^{-8}$

(a) First four terms of $\cos x$ A1

Standard series: $\cos x = 1 - \tfrac{x^{2}}{2!} + \tfrac{x^{4}}{4!} - \tfrac{x^{6}}{6!} + \cdots$. Evaluate denominators: $$ \cos x \;\approx\; 1 \;-\; \tfrac{x^{2}}{2} \;+\; \tfrac{x^{4}}{24} \;-\; \tfrac{x^{6}}{720}. $$

(b) Substitute $x = 0.5$ M1·A1·A1

$1 = 1.000000$.
$\tfrac{(0.5)^{2}}{2} = \tfrac{0.25}{2} = 0.125000$.
$\tfrac{(0.5)^{4}}{24} = \tfrac{0.0625}{24} \approx 0.002604$.
$\tfrac{(0.5)^{6}}{720} = \tfrac{0.015625}{720} \approx 0.0000217$.

(c) Sum to $4$ d.p. A1

$1 - 0.125000 + 0.002604 - 0.0000217 \approx 0.877583$. Rounded to $4$ d.p.: $\cos(0.5) \approx \boxed{0.8776}$.

(d) Compare with GDC A1·A1

GDC value: $\cos(0.5) = 0.87758256\ldots \approx 0.8776$ to $4$ d.p. Using the unrounded partial sum $S_{4} = 0.87758246527\ldots$, the absolute error of the $4$-term approximation is $$ |\,S_{4} - \cos(0.5)\,| \;=\; |\,0.87758246527 - 0.87758256189\,| \;\approx\; 9.66 \times 10^{-8}. $$ So the approximation matches the GDC to $4$ d.p. as required, and in fact to $7$ d.p. once the partial sum is kept unrounded.

Why does a $4$-term truncation crush $\cos(0.5)$? The Maclaurin series for $\cos x$ is alternating, with terms whose absolute values decrease monotonically once $|x| < 1$. By the alternating-series remainder theorem, the truncation error is bounded by the magnitude of the first omitted term: $\bigl|\tfrac{(0.5)^{8}}{8!}\bigr| = \tfrac{0.00390625}{40320} \approx 9.69 \times 10^{-8}$. The actual error ($\approx 9.66 \times 10^{-8}$) sits right at this bound — the next omitted term $\bigl|\tfrac{(0.5)^{10}}{10!}\bigr| \approx 2.7 \times 10^{-10}$ is two orders of magnitude smaller, so it barely shifts the result. Two heuristics: (i) for $|x| \le 1$, each new term of $\cos x$ or $\sin x$ shrinks by a factor $\sim x^{2}/n(n-1) \le 1/(n(n-1))$, so the $n$-th term is roughly $1/n!$; (ii) "trustworthy decimal places" $\approx$ "first negligible term has that many leading zeros." For $x = 0.5$, four terms already give $7$ decimal places — a single Paper 2 partial sum.

(a) $\cos x$ 前四项 A1

标准级数 $\cos x = 1 - \tfrac{x^{2}}{2!} + \tfrac{x^{4}}{4!} - \tfrac{x^{6}}{6!} + \cdots$。化分母： $$ \cos x \;\approx\; 1 \;-\; \tfrac{x^{2}}{2} \;+\; \tfrac{x^{4}}{24} \;-\; \tfrac{x^{6}}{720}. $$

(b) 代 $x = 0.5$ M1·A1·A1

$1 = 1.000000$。
$\tfrac{(0.5)^{2}}{2} = 0.125000$。
$\tfrac{(0.5)^{4}}{24} \approx 0.002604$。
$\tfrac{(0.5)^{6}}{720} \approx 0.0000217$。

(c) 求和至 $4$ 位小数 A1

$1 - 0.125000 + 0.002604 - 0.0000217 \approx 0.877583$。四舍五入到 $4$ 位：$\cos(0.5) \approx \boxed{0.8776}$。

(d) 与 GDC 比较 A1·A1

GDC：$\cos(0.5) = 0.87758256\ldots \approx 0.8776$（$4$ 位）。取未四舍的部分和 $S_{4} = 0.87758246527\ldots$，四项近似的绝对误差为 $$ |\,S_{4} - \cos(0.5)\,| \;=\; |\,0.87758246527 - 0.87758256189\,| \;\approx\; 9.66 \times 10^{-8}. $$ 四项近似与 GDC 在 $4$ 位上一致，未四舍部分和可达 $7$ 位。

为何 $4$ 项就把 $\cos(0.5)$ 估算到这么准？$\cos x$ 麦克劳林级数为交错级数，且 $|x| < 1$ 时各项绝对值单调递减。由交错级数余项定理，截断误差不超过首个被略项的绝对值：$\bigl|\tfrac{(0.5)^{8}}{8!}\bigr| \approx 9.69 \times 10^{-8}$。实际误差（$\approx 9.66 \times 10^{-8}$）正贴近这个上界——因为下一个被略项 $\bigl|\tfrac{(0.5)^{10}}{10!}\bigr| \approx 2.7 \times 10^{-10}$ 比之小两个数量级，几乎不动结果。两条经验：(i) $|x| \le 1$ 时 $\cos x$、$\sin x$ 每加一项缩小约 $1/(n(n-1))$，第 $n$ 项约为 $1/n!$；(ii) "可信小数位数"约等于"首个被略项前导零的个数"。$x = 0.5$ 处四项即得 $7$ 位小数——一行 Paper 2 部分和即可。

Q7HARDPaper 2AHL 5.18 Limit via Series Expansion (HL)[10 marks]

$L = \displaystyle\lim_{x \to 0} \dfrac{\sin x - x}{x^{3}}$. Series method (a)–(c); verify with L'Hopital $\times 3$ (d).$L = \displaystyle\lim_{x \to 0} \dfrac{\sin x - x}{x^{3}}$。级数法 (a)–(c)；用三次洛必达验证 (d)。

Answer:答案： $L = -\dfrac{1}{6}$

(a) Series for $\sin x$ A1·A1

Standard: $\sin x = x - \tfrac{x^{3}}{3!} + \tfrac{x^{5}}{5!} - \cdots = x - \tfrac{x^{3}}{6} + \tfrac{x^{5}}{120} - \cdots$. First three non-zero terms: $x - \tfrac{x^{3}}{6} + \tfrac{x^{5}}{120}$.

(b) Subtract $x$, divide by $x^{3}$ M1·A1·A1

$$ \sin x - x \;=\; -\tfrac{x^{3}}{6} + \tfrac{x^{5}}{120} - \cdots. $$ $$ \frac{\sin x - x}{x^{3}} \;=\; -\tfrac{1}{6} + \tfrac{x^{2}}{120} - \cdots. $$

(c) Take $x \to 0$ M1·A1

Every term with positive power of $x$ vanishes: $$ L \;=\; \lim_{x \to 0}\!\left( -\tfrac{1}{6} + \tfrac{x^{2}}{120} - \tfrac{x^{4}}{5040} + \cdots \right) \;=\; -\tfrac{1}{6}. $$

(d) Verify with L'Hopital $\times 3$ M1·A1·A1

Direct substitution: $\sin 0 - 0 = 0$, $0^{3} = 0$. Form $0/0$. $$ L = \lim_{x \to 0} \frac{\sin x - x}{x^{3}} \stackrel{\text{LH}}{=} \lim_{x \to 0} \frac{\cos x - 1}{3 x^{2}}. $$ At $x = 0$: $\cos 0 - 1 = 0$, $3 \cdot 0^{2} = 0$. Form $0/0$ again. $$ \stackrel{\text{LH}}{=} \lim_{x \to 0} \frac{-\sin x}{6 x}. $$ At $x = 0$: $-\sin 0 = 0$, $6 \cdot 0 = 0$. Form $0/0$ once more. $$ \stackrel{\text{LH}}{=} \lim_{x \to 0} \frac{-\cos x}{6} \;=\; \frac{-1}{6} \;=\; -\tfrac{1}{6}. \;\checkmark $$

Series replace iterated L'Hopital — and usually win. The series method does all three L'Hopital applications in one line. The rule of thumb: when a limit at $0$ has the form $\dfrac{f(x)}{x^{k}}$ with $f(0) = f'(0) = \cdots = f^{(k-1)}(0) = 0$, the limit equals $\dfrac{f^{(k)}(0)}{k!}$ — exactly the Maclaurin coefficient of $x^{k}$ in $f$. Here $f(x) = \sin x - x$ has Maclaurin coefficient $-\tfrac{1}{6}$ at $x^{3}$, so the limit is $-\tfrac{1}{6}$. Once you internalise this, "find the lowest non-vanishing Maclaurin coefficient" becomes a single mental step. Compare with L'Hopital: $3$ differentiations, $3$ form-checks, $9$ chances to slip a sign. The series method is faster, less error-prone, and works on indeterminate forms that L'Hopital cannot crack cleanly (e.g. limits involving $x^{x}$ or $(\sin x)^{x}$ benefit from $\ln$ then series, not iterated L'Hopital).

(a) $\sin x$ 级数 A1·A1

标准：$\sin x = x - \tfrac{x^{3}}{6} + \tfrac{x^{5}}{120} - \cdots$。前三个非零项：$x - \tfrac{x^{3}}{6} + \tfrac{x^{5}}{120}$。

(b) 减 $x$ 后除以 $x^{3}$ M1·A1·A1

$$ \sin x - x \;=\; -\tfrac{x^{3}}{6} + \tfrac{x^{5}}{120} - \cdots,\qquad \frac{\sin x - x}{x^{3}} \;=\; -\tfrac{1}{6} + \tfrac{x^{2}}{120} - \cdots. $$

(c) 取 $x \to 0$ M1·A1

$x$ 正幂项均消失：$L = -\tfrac{1}{6}$。

(d) 三次洛必达验证 M1·A1·A1

代入 $0$：$0/0$。 $$ L \stackrel{\text{LH}}{=} \lim_{x \to 0} \frac{\cos x - 1}{3 x^{2}} \stackrel{0/0}{=} \lim_{x \to 0} \frac{-\sin x}{6 x} \stackrel{0/0}{=} \lim_{x \to 0} \frac{-\cos x}{6} \;=\; -\tfrac{1}{6}. \;\checkmark $$

级数替代多次洛必达——通常更胜。级数方法把三次洛必达浓缩为一行。经验法则：在 $0$ 处，若极限形如 $\dfrac{f(x)}{x^{k}}$ 且 $f(0) = f'(0) = \cdots = f^{(k-1)}(0) = 0$，则极限 $= \dfrac{f^{(k)}(0)}{k!}$——恰为 $f$ 的麦克劳林级数中 $x^{k}$ 的系数。本题 $f(x) = \sin x - x$ 的 $x^{3}$ 系数为 $-\tfrac{1}{6}$，故 $L = -\tfrac{1}{6}$。内化后，"找最低非零麦克劳林系数"就是一个心算步骤。对比洛必达：$3$ 次求导、$3$ 次形式核查、$9$ 次符号机会。级数法更快、更稳，并能处理洛必达难以干净处理的不定型（如 $x^{x}$、$(\sin x)^{x}$ 须先取 $\ln$ 再用级数，而非反复洛必达）。

Paper 3 · Worked Solutions第三卷 · 详细解析

Q8HARDPaper 3AHL 5.18 Convergence Bound + Partial Sums (HL)[15 marks]

$S_{n} = \sum_{k=0}^{n} \tfrac{1}{k!}$ approximates $e$. Find the smallest $n$ with $|S_{n} - e| < 0.001$ using Lagrange's remainder bound, then verify.$S_{n} = \sum_{k=0}^{n} \tfrac{1}{k!}$ 近似 $e$。用拉格朗日余项界求使 $|S_{n} - e| < 0.001$ 的最小 $n$，并核验。

Answers:答案： (c) $n = 6$ · (d) $S_{6} = \tfrac{1957}{720}$ · (e) $|S_{6} - e| \approx 2.26 \times 10^{-4}$

(a) Series for $e^{x}$, $R = \infty$ A1·A1

$e^{x} = \sum_{k=0}^{\infty} \tfrac{x^{k}}{k!}$. Radius of convergence: by the ratio test, $\lim_{k \to \infty} \tfrac{|x|^{k+1}/(k+1)!}{|x|^{k}/k!} = \lim_{k \to \infty} \tfrac{|x|}{k+1} = 0 < 1$ for every $x$. Hence $R = \infty$ and the interval of convergence is $\mathbb{R}$. In particular $x = 1$ lies inside, and $S_{n} \to e$ as $n \to \infty$.

(b) Lagrange bound on $[0, 1]$ M1·A1·A1

Lagrange's form of the remainder: there exists $\xi \in (0, 1)$ with $$ R_{n}(1) \;=\; \frac{f^{(n+1)}(\xi)}{(n+1)!}\, (1 - 0)^{n+1} \;=\; \frac{e^{\xi}}{(n+1)!}. $$ Since $0 < \xi < 1$ and $e^{x}$ is increasing, $0 < e^{\xi} < e^{1} = e$. Hence $$ |R_{n}(1)| \;\le\; \frac{e}{(n+1)!}. \;\checkmark $$

(c) Smallest $n$ with bound $< 0.001$ M1·A1·A1

Need $\dfrac{e}{(n+1)!} < 0.001 = 10^{-3}$. Using $e < 2.72$: $(n+1)! > 2720$. Tabulate:

$n = 5$: $(n+1)! = 6! = 720 < 2720$. Bound fails.
$n = 6$: $(n+1)! = 7! = 5040 > 2720$. Bound holds: $\tfrac{2.72}{5040} \approx 5.4 \times 10^{-4} < 10^{-3}$. $\checkmark$

Smallest $n$ from the bound: $\boxed{n = 6}$.

(d) Compute $S_{6}$ as a fraction M1·A1·A1

$S_{6} = 1 + 1 + \tfrac{1}{2} + \tfrac{1}{6} + \tfrac{1}{24} + \tfrac{1}{120} + \tfrac{1}{720}$. Common denominator $720$: $$ S_{6} \;=\; \frac{720 + 720 + 360 + 120 + 30 + 6 + 1}{720} \;=\; \frac{1957}{720}. $$ Check $\gcd(1957, 720)$: $1957 = 2 \cdot 720 + 517$; $\gcd(720, 517) = \gcd(517, 203) = \gcd(203, 111) = \gcd(111, 92) = \gcd(92, 19) = \gcd(19, 16) = \gcd(16, 3) = \gcd(3, 1) = 1$. So $\tfrac{1957}{720}$ is in lowest terms.

(e) Compare with GDC value of $e$ A1·A1

GDC: $e = 2.7182818\ldots$. Decimal of $S_{6}$: $\tfrac{1957}{720} = 2.7180555\ldots$. Actual error: $$ |S_{6} - e| \;=\; |\,2.7180555\ldots - 2.7182818\ldots\,| \;\approx\; 2.26 \times 10^{-4}. $$ The Lagrange bound predicted $\le 5.4 \times 10^{-4}$. The actual error is roughly half the bound — Lagrange is correct but not tight; the truth sits in the middle.

(f) Confirm $n - 1 = 5$ fails M1·A1

$S_{5} = 1 + 1 + \tfrac{1}{2} + \tfrac{1}{6} + \tfrac{1}{24} + \tfrac{1}{120} = \tfrac{120 + 120 + 60 + 20 + 5 + 1}{120} = \tfrac{326}{120} = \tfrac{163}{60} \approx 2.71666\ldots$. $$ |S_{5} - e| \;\approx\; |\,2.71667 - 2.71828\,| \;\approx\; 1.62 \times 10^{-3} \;>\; 10^{-3}. $$ So $n = 5$ does not meet the tolerance, and $n = 6$ is indeed the smallest non-negative integer with $|S_{n} - e| < 0.001$.

Lagrange bound vs. actual error — and what "smallest $n$" really means. There are two distinct "smallest $n$" answers floating around: (i) the smallest $n$ such that the Lagrange bound guarantees the tolerance (here $n = 6$, with bound $5.4 \times 10^{-4}$); (ii) the smallest $n$ such that the actual error meets the tolerance (here also $n = 6$, since $n = 5$ gives $1.62 \times 10^{-3} > 10^{-3}$). In this problem the two coincide, but in general the bound is conservative — you may meet the tolerance with fewer terms than the Lagrange estimate predicts. The IB question specifies "bound guarantees" in (c) and asks for actual verification in (e)/(f). Always read the question carefully: if it says "use Lagrange's bound to find $n$" you answer with the bound; if it says "the smallest $n$ with $|R_{n}| < \epsilon$" you must verify actual error and the next-smaller $n$ failing. Three guardrails: (1) the Lagrange bound on $[0, b]$ for $e^{x}$ is $\dfrac{e^{b}}{(n+1)!}\, b^{n+1}$ — both factors matter; (2) factorials in the denominator dominate any power growth, so doubling $n$ typically chops the error by orders of magnitude; (3) for alternating series ($\sin$, $\cos$, $\ln(1+x)$, $\arctan x$ near interior points), the alternating-series bound is often tighter than Lagrange — pick whichever the problem allows.

(a) $e^{x}$ 级数，$R = \infty$ A1·A1

$e^{x} = \sum_{k=0}^{\infty} \tfrac{x^{k}}{k!}$。由比值判别法：$\lim_{k \to \infty} \tfrac{|x|}{k+1} = 0 < 1$（对每个 $x$），故 $R = \infty$，收敛区间 $\mathbb{R}$；$x = 1$ 在内，$S_{n} \to e$。

(b) $[0, 1]$ 上的拉格朗日界 M1·A1·A1

拉格朗日余项形式：存在 $\xi \in (0, 1)$ 使 $$ R_{n}(1) \;=\; \frac{f^{(n+1)}(\xi)}{(n+1)!}\, (1 - 0)^{n+1} \;=\; \frac{e^{\xi}}{(n+1)!}. $$ 由 $0 < \xi < 1$、$e^{x}$ 递增：$0 < e^{\xi} < e^{1} = e$。故 $$ |R_{n}(1)| \;\le\; \frac{e}{(n+1)!}. \;\checkmark $$

(c) 使界 $< 0.001$ 的最小 $n$ M1·A1·A1

需 $\dfrac{e}{(n+1)!} < 10^{-3}$。由 $e < 2.72$：$(n+1)! > 2720$。

$n = 5$：$6! = 720 < 2720$。不满足。
$n = 6$：$7! = 5040 > 2720$。$\tfrac{2.72}{5040} \approx 5.4 \times 10^{-4} < 10^{-3}$。$\checkmark$

由界给出的最小 $n$：$\boxed{n = 6}$。

(d) 把 $S_{6}$ 写为分数 M1·A1·A1

$S_{6} = 1 + 1 + \tfrac{1}{2} + \tfrac{1}{6} + \tfrac{1}{24} + \tfrac{1}{120} + \tfrac{1}{720}$。公分母 $720$： $$ S_{6} \;=\; \frac{720 + 720 + 360 + 120 + 30 + 6 + 1}{720} \;=\; \frac{1957}{720}. $$ 验 $\gcd(1957, 720) = 1$（辗转相除），故 $\tfrac{1957}{720}$ 已最简。

(e) 与 GDC 的 $e$ 比较 A1·A1

GDC：$e = 2.7182818\ldots$；$\tfrac{1957}{720} = 2.7180555\ldots$。实际误差： $$ |S_{6} - e| \;\approx\; 2.26 \times 10^{-4}. $$ 拉格朗日界预测 $\le 5.4 \times 10^{-4}$，实际约为界的一半——界正确但不紧，真值居中。

(f) 验证 $n - 1 = 5$ 不满足 M1·A1

$S_{5} = \tfrac{163}{60} \approx 2.71667$；$|S_{5} - e| \approx 1.62 \times 10^{-3} > 10^{-3}$。故 $n = 5$ 不满足，$n = 6$ 确为最小。

拉格朗日界 vs. 实际误差——"最小 $n$" 究竟指哪个？有两种"最小 $n$"：(i) 使拉格朗日界保证容差的最小 $n$（本题 $n = 6$，界为 $5.4 \times 10^{-4}$）；(ii) 使实际误差满足容差的最小 $n$（本题也是 $n = 6$，因为 $n = 5$ 实际误差 $1.62 \times 10^{-3} > 10^{-3}$）。本题两者一致，但一般界是保守的——可能用更少的项就已满足容差。IB 此题在 (c) 中要求"用拉格朗日界保证"，在 (e)/(f) 中要求实际验证。务必看清题目：写"用拉格朗日界求 $n$"就按界回答；写"满足 $|R_{n}| < \epsilon$ 的最小 $n$"就要验实际误差且少一项失败。三条护栏：(1) $e^{x}$ 在 $[0, b]$ 上拉格朗日界为 $\dfrac{e^{b}}{(n+1)!}\, b^{n+1}$，两个因子都重要；(2) 分母的阶乘压倒任意幂增长，$n$ 加倍误差通常下降数量级；(3) 交错级数（$\sin$、$\cos$、$\ln(1+x)$、$\arctan x$ 在内点附近）的交错界常比拉格朗日更紧——题目允许时择优用。