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Unit B7 · Solutions第B7单元 · 解答

Power, Taylor, and Maclaurin Series · Solutions

Companion to the University-Style Practice Set大学风格练习题配套解答

MEDIUM HARD CORE PROOF APPLIED

Sections 1 to 7: radius and interval of convergence, power series representations, differentiation and integration of power series, Taylor and Maclaurin series, key Maclaurin library, Taylor remainder, applications第 1 至 7 节：收敛半径与收敛区间，幂级数表示，幂级数的微分与积分，泰勒级数与麦克劳林级数，常用麦克劳林公式库，泰勒余项，应用CALC II

PART I  ·  CORE TECHNIQUES核心技巧Computational fluency · 28 marks计算熟练度 · 28分

Worked Solutions详细解答

Q1MEDIUMCOREratio test, radius and interval of convergence比值审敛法，收敛半径与收敛区间[8 marks]

(a) Find the radius and interval of convergence of $\sum_{n=1}^{\infty}\frac{(-1)^{n}x^{n}}{n\cdot 3^{n}}$; (b) same for $\sum_{n=0}^{\infty}\frac{(x-2)^{n}}{n^{2}+1}$. Check both endpoints in each case.(a) 求 $\sum_{n=1}^{\infty}\frac{(-1)^{n}x^{n}}{n\cdot 3^{n}}$ 的收敛半径和收敛区间；(b) 对 $\sum_{n=0}^{\infty}\frac{(x-2)^{n}}{n^{2}+1}$ 做同样的分析，两题均需检验两个端点。

Answers:答案：  (a) $R=3$, interval区间 $(-3,\,3]$  ·  (b) $R=1$, interval区间 $[1,\,3]$

(a) Ratio test for $\sum \frac{(-1)^{n}x^{n}}{n\cdot 3^{n}}$(a) 对 $\sum \frac{(-1)^{n}x^{n}}{n\cdot 3^{n}}$ 用比值审敛法 M1·A1·A1·A1

Let $a_{n}=\dfrac{|x|^{n}}{n\cdot 3^{n}}$. Form the ratio: (M1)令 $a_{n}=\dfrac{|x|^{n}}{n\cdot 3^{n}}$，构造比值：(M1)

$$ \frac{a_{n+1}}{a_{n}}=\frac{|x|^{n+1}}{(n+1)\cdot 3^{n+1}}\cdot\frac{n\cdot 3^{n}}{|x|^{n}}=\frac{n}{n+1}\cdot\frac{|x|}{3}\longrightarrow \frac{|x|}{3}\quad\text{as }n\to\infty. $$

The series converges when $|x|/3<1$, i.e. $|x|<3$, so $R=3$. (A1)当 $|x|/3<1$，即 $|x|<3$ 时级数收敛，故 $R=3$。(A1)

Endpoint $x=3$:端点 $x=3$： the series becomes $\sum_{n=1}^{\infty}\dfrac{(-1)^{n}}{n}$, which converges by the alternating series test (terms $1/n$ decrease monotonically to $0$). (A1)级数变为 $\sum_{n=1}^{\infty}\dfrac{(-1)^{n}}{n}$，由交错级数审敛法收敛（各项 $1/n$ 单调递减趋于 $0$）。(A1)

Endpoint $x=-3$:端点 $x=-3$： the series becomes $\sum_{n=1}^{\infty}\dfrac{(-1)^{n}(-1)^{n}}{n}=\sum_{n=1}^{\infty}\dfrac{1}{n}$, the divergent harmonic series. (A1)级数变为 $\sum_{n=1}^{\infty}\dfrac{(-1)^{n}(-1)^{n}}{n}=\sum_{n=1}^{\infty}\dfrac{1}{n}$，即发散的调和级数。(A1)

Interval of convergence: $(-3,\,3]$.收敛区间：$(-3,\,3]$。

(b) Ratio test for $\sum \frac{(x-2)^{n}}{n^{2}+1}$(b) 对 $\sum \frac{(x-2)^{n}}{n^{2}+1}$ 用比值审敛法 M1·A1·A1·A1

Let $u=x-2$. The ratio of successive terms satisfies (M1)令 $u=x-2$，相邻两项之比满足 (M1)

$$ \frac{|u|^{n+1}/((n+1)^{2}+1)}{|u|^{n}/(n^{2}+1)}\approx |u|\cdot\frac{n^{2}+1}{(n+1)^{2}+1}\longrightarrow |u|\quad\text{as }n\to\infty. $$

So $R=1$, meaning convergence for $|x-2|<1$, i.e. $1故 $R=1$，即当 $|x-2|<1$，即 $1

Endpoint $x=1$ ($u=-1$):端点 $x=1$（$u=-1$）： the series becomes $\sum_{n=0}^{\infty}\dfrac{(-1)^{n}}{n^{2}+1}$, which converges absolutely since $\sum\dfrac{1}{n^{2}+1}$ converges by comparison with $\sum 1/n^{2}$ (p-series, $p=2>1$). (A1)级数变为 $\sum_{n=0}^{\infty}\dfrac{(-1)^{n}}{n^{2}+1}$，由于 $\sum\dfrac{1}{n^{2}+1}$ 通过与 $\sum 1/n^{2}$（$p$级数，$p=2>1$）比较可知收敛，故绝对收敛。(A1)

Endpoint $x=3$ ($u=1$):端点 $x=3$（$u=1$）： the series becomes $\sum_{n=0}^{\infty}\dfrac{1}{n^{2}+1}$, which converges by the same comparison. (A1)级数变为 $\sum_{n=0}^{\infty}\dfrac{1}{n^{2}+1}$，由相同比较法收敛。(A1)

Interval of convergence: $[1,\,3]$.收敛区间：$[1,\,3]$。

Insight. Always test both endpoints after the ratio test; the answer changes case by case. In (a) the alternating sign at $x=3$ saves convergence, while $x=-3$ kills the sign and exposes the harmonic series. In (b) the $n^{2}$ denominator gives absolute convergence at both ends without any alternating trick. Never skip the endpoint check, and always name the test you invoke there.用比值审敛法后务必检验两个端点，结论因情况而异。在 (a) 中，$x=3$ 处的交错符号使级数收敛，而 $x=-3$ 处符号抵消暴露出调和级数。在 (b) 中，分母的 $n^{2}$ 使两端均绝对收敛，无需交错技巧。端点检验不可省略，且须写明所用判别法的名称。
Q2HARDCOREratio test, factorial growth, endpoint classification比值审敛法，阶乘增长，端点分类[10 marks]

(a) Radius and interval for $\sum_{n=0}^{\infty}\frac{n!\,(x-1)^{n}}{5^{n}}$; (b) radius and interval for $\sum_{n=1}^{\infty}\frac{(2x+1)^{n}}{n\cdot 4^{n}}$. Name every endpoint test.(a) 求 $\sum_{n=0}^{\infty}\frac{n!\,(x-1)^{n}}{5^{n}}$ 的收敛半径和收敛区间；(b) 求 $\sum_{n=1}^{\infty}\frac{(2x+1)^{n}}{n\cdot 4^{n}}$ 的收敛半径和收敛区间，每个端点均需写明判别法名称。

Answers:答案：  (a) $R=0$, converges only at $x=1$仅在 $x=1$ 处收敛  ·  (b) $R=2$ (in $x$关于 $x$), interval区间 $\bigl[-\tfrac{5}{2},\,\tfrac{3}{2}\bigr)$

(a) Factorial growth forces $R=0$(a) 阶乘增长迫使 $R=0$ M1·A1·R1·A1

Form the ratio of successive terms: (M1)构造相邻两项之比：(M1)

$$ \frac{(n+1)!\,|x-1|^{n+1}/5^{n+1}}{n!\,|x-1|^{n}/5^{n}}=(n+1)\cdot\frac{|x-1|}{5}\longrightarrow\infty\quad\text{for any }x\ne 1. $$

Since the ratio diverges to infinity for every $x\ne 1$, the series diverges everywhere except $x=1$. (A1)由于对所有 $x\ne 1$，比值趋于无穷，级数在除 $x=1$ 外的所有点均发散。(A1)

At $x=1$ every term is $0$, so the series converges trivially. (R1)在 $x=1$ 处每项均为 $0$，故级数显然收敛。(R1)

Radius of convergence $R=0$; the interval of convergence is the single point $\{1\}$. (A1)收敛半径 $R=0$；收敛区间为单点 $\{1\}$。(A1)

(b) Rewrite in terms of $u=2x+1$, then ratio test(b) 换元 $u=2x+1$ 后用比值审敛法 M1·A1·A1·A1·A1·A1

Write $u=2x+1$. The series is $\sum_{n=1}^{\infty}\dfrac{u^{n}}{n\cdot 4^{n}}$. The ratio of consecutive terms: (M1)令 $u=2x+1$，级数化为 $\sum_{n=1}^{\infty}\dfrac{u^{n}}{n\cdot 4^{n}}$，相邻两项之比为：(M1)

$$ \frac{|u|^{n+1}/((n+1)\cdot 4^{n+1})}{|u|^{n}/(n\cdot 4^{n})}=\frac{n}{n+1}\cdot\frac{|u|}{4}\longrightarrow \frac{|u|}{4}. $$

Convergence requires $|u|/4<1$, i.e. $|2x+1|<4$. Solving: $-5<2x+1<4$, so $-\tfrac{5}{2}收敛要求 $|u|/4<1$，即 $|2x+1|<4$，解得 $-5<2x+1<4$，故 $-\tfrac{5}{2}

Endpoint $x=3/2$ ($u=4$):端点 $x=3/2$（$u=4$）： the series becomes $\sum_{n=1}^{\infty}\dfrac{4^{n}}{n\cdot 4^{n}}=\sum_{n=1}^{\infty}\dfrac{1}{n}$, the divergent harmonic series (p-series, $p=1$). (A1)级数变为 $\sum_{n=1}^{\infty}\dfrac{4^{n}}{n\cdot 4^{n}}=\sum_{n=1}^{\infty}\dfrac{1}{n}$，即发散的调和级数（$p$级数，$p=1$）。(A1)

Endpoint $x=-5/2$ ($u=-4$):端点 $x=-5/2$（$u=-4$）： the series becomes $\sum_{n=1}^{\infty}\dfrac{(-4)^{n}}{n\cdot 4^{n}}=\sum_{n=1}^{\infty}\dfrac{(-1)^{n}}{n}$, which converges by the alternating series test (terms $1/n$ decrease monotonically to $0$). (A1)级数变为 $\sum_{n=1}^{\infty}\dfrac{(-4)^{n}}{n\cdot 4^{n}}=\sum_{n=1}^{\infty}\dfrac{(-1)^{n}}{n}$，由交错级数审敛法收敛（各项 $1/n$ 单调递减趋于 $0$）。(A1)

Interval of convergence: $\bigl[-\tfrac{5}{2},\,\tfrac{3}{2}\bigr)$. (A1)收敛区间：$\bigl[-\tfrac{5}{2},\,\tfrac{3}{2}\bigr)$。(A1)

Insight. Whenever the series is in powers of $(ax+b)$ rather than $x$, translate to $u=ax+b$ first, find the interval for $u$, then back-substitute for $x$. The radius in $x$ is $R_{u}/|a|$; here $4/2=2$. The endpoint story is identical to Q1(a): the point where $u=4$ exposes the harmonic series (diverges) and where $u=-4$ gives the alternating harmonic series (converges).当级数以 $(ax+b)$ 的幂次给出而非 $x$ 时，先换元 $u=ax+b$，求 $u$ 的收敛区间，再回代为 $x$。关于 $x$ 的收敛半径为 $R_{u}/|a|$，此处为 $4/2=2$。端点情形与第1题(a)相同：$u=4$ 暴露调和级数（发散），$u=-4$ 给出交错调和级数（收敛）。
Q3MEDIUMCOREbuilding new series by substitution and integration通过换元与积分构造新级数[10 marks]

Starting from $\frac{1}{1-x}=\sum_{n=0}^{\infty}x^{n}$ ($|x|<1$): (a) derive $\frac{1}{1+x^{2}}$ as a series; (b) integrate to get the Maclaurin series for $\arctan x$; (c) evaluate at $x=1$ to express $\pi/4$ as a series.从 $\frac{1}{1-x}=\sum_{n=0}^{\infty}x^{n}$（$|x|<1$）出发：(a) 导出 $\frac{1}{1+x^{2}}$ 的级数；(b) 积分得到 $\arctan x$ 的麦克劳林级数；(c) 代入 $x=1$ 将 $\pi/4$ 表示为级数。

Answers:答案：  (a) $\sum_{n=0}^{\infty}(-1)^{n}x^{2n}$, $|x|<1$  ·  (b) $\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{2n+1}$, $[-1,1]$  ·  (c) $\tfrac{\pi}{4}=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}$, alternating series test交错级数审敛法

(a) Substitute $x\mapsto -x^{2}$(a) 令 $x\mapsto -x^{2}$ M1·A1·A1

Replace $x$ by $-x^{2}$ in the geometric series: (M1)在等比级数中以 $-x^{2}$ 替换 $x$：(M1)

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

The substitution requires $|-x^{2}|<1$, i.e. $|x|<1$. (A1) The interval of convergence is $(-1,1)$. (A1)换元要求 $|-x^{2}|<1$，即 $|x|<1$。(A1) 收敛区间为 $(-1,1)$。(A1)

(b) Integrate term by term(b) 逐项积分 M1·A1·M1·A1

Since the series converges on an open interval, it may be integrated term by term: (M1)由于级数在开区间上收敛，可以逐项积分：(M1)

$$ \arctan x=\int_{0}^{x}\frac{dt}{1+t^{2}}=\int_{0}^{x}\sum_{n=0}^{\infty}(-1)^{n}t^{2n}\,dt=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{2n+1}. $$

The constant of integration is $0$ because $\arctan 0=0$. (A1)积分常数为 $0$，因为 $\arctan 0=0$。(A1)

The radius of convergence remains $1$. (M1) At $x=1$ the series $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}$ converges by the alternating series test (terms $\frac{1}{2n+1}$ decrease monotonically to $0$), so the interval extends to include $x=1$. At $x=-1$ the series becomes $\sum_{n=0}^{\infty}\frac{(-1)^{n}(-1)^{2n+1}}{2n+1}=-\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}$, which also converges by the alternating series test. Interval of convergence: $[-1,1]$ for the Maclaurin series of $\arctan x$. (A1)收敛半径仍为 $1$。(M1) 在 $x=1$ 处，级数 $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}$ 由交错级数审敛法收敛（各项 $\frac{1}{2n+1}$ 单调递减趋于 $0$），故区间扩展至包含 $x=1$。在 $x=-1$ 处，级数变为 $-\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}$，同样由交错级数审敛法收敛。$\arctan x$ 的麦克劳林级数收敛区间：$[-1,1]$。(A1)

(c) Leibniz formula for $\pi$(c) $\pi$ 的莱布尼茨公式 M1·A1·R1

Since the series converges at $x=1$ and $\arctan 1=\pi/4$, substitute $x=1$: (M1)由于级数在 $x=1$ 处收敛，且 $\arctan 1=\pi/4$，代入 $x=1$：(M1)

$$ \frac{\pi}{4}=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots $$

(A1) Convergence at the endpoint is guaranteed by the alternating series test: the terms $\frac{1}{2n+1}$ are positive, strictly decreasing, and tend to $0$. (R1)(A1) 端点处的收敛由交错级数审敛法保证：各项 $\frac{1}{2n+1}$ 为正，严格递减，趋于 $0$。(R1)

Insight. Substitution into a known series is far faster than computing Taylor coefficients from scratch. The key moves are: (1) identify what must replace $x$ to get the target function, (2) keep track of the resulting restriction on $x$ (here $|-x^{2}|<1$ gives $|x|<1$), (3) integrate or differentiate term by term. At integration, never forget to determine the constant of integration from an initial value, and always re-examine the endpoint(s) separately since integration can enlarge the interval.对已知级数换元比从头计算泰勒系数快得多。关键步骤：(1) 确定用什么替换 $x$ 可得目标函数，(2) 追踪对 $x$ 的限制（此处 $|-x^{2}|<1$ 给出 $|x|<1$），(3) 逐项积分或微分。积分时切勿忘记用初值确定积分常数，且积分后可能扩大收敛区间，须单独检验端点。
PART II  ·  DEFINITIONS AND PROOF定义与证明Rigorous arguments · 26 marks严格论证 · 26分

Worked Solutions详细解答

Q4HARDPROOFMaclaurin series from the definition; interval of convergence proof从定义出发的麦克劳林级数；收敛区间证明[10 marks]

For $f(x)=\ln(1+x)$: (a) compute the first five values $f^{(n)}(0)$ and write the Maclaurin series; (b) show $R=1$ by the ratio test; (c) test $x=1$; (d) test $x=-1$ and state the interval.对 $f(x)=\ln(1+x)$：(a) 计算前五个值 $f^{(n)}(0)$ 并写出麦克劳林级数；(b) 用比值审敛法证明 $R=1$；(c) 检验 $x=1$；(d) 检验 $x=-1$ 并写出收敛区间。

Answers:答案：  (a) $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}x^{n}$  ·  (b) $R=1$  ·  (c) converges (alternating series test)收敛（交错级数审敛法）  ·  (d) diverges; interval $(-1,1]$发散；收敛区间 $(-1,1]$

(a) Computing derivatives at $0$(a) 计算在 $0$ 处的各阶导数 M1·M1·A1·A1·A1

Differentiate repeatedly: (M1)反复求导：(M1)

$f(x)=\ln(1+x)$, $f(0)=0$.

$f'(x)=(1+x)^{-1}$, $f'(0)=1$.

$f''(x)=-(1+x)^{-2}$, $f''(0)=-1$.

$f'''(x)=2(1+x)^{-3}$, $f'''(0)=2$.

$f^{(4)}(x)=-6(1+x)^{-4}$, $f^{(4)}(0)=-6$. (M1)

The pattern gives $f^{(n)}(0)=(-1)^{n-1}(n-1)!$ for $n\ge 1$. (A1) The Maclaurin coefficient is $c_{n}=f^{(n)}(0)/n!=(-1)^{n-1}/n$. (A1) Therefore规律给出 $f^{(n)}(0)=(-1)^{n-1}(n-1)!$（$n\ge 1$）。(A1) 麦克劳林系数为 $c_{n}=f^{(n)}(0)/n!=(-1)^{n-1}/n$。(A1) 因此

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

(A1)

(b) Ratio test for the radius(b) 用比值审敛法求收敛半径 M1·A1

With $a_{n}=\dfrac{|x|^{n}}{n}$: (M1)令 $a_{n}=\dfrac{|x|^{n}}{n}$：(M1)

$$ \frac{a_{n+1}}{a_{n}}=\frac{|x|^{n+1}}{n+1}\cdot\frac{n}{|x|^{n}}=\frac{n}{n+1}\,|x|\longrightarrow |x|. $$

The series converges for $|x|<1$, so $R=1$. (A1)当 $|x|<1$ 时级数收敛，故 $R=1$。(A1)

(c) Endpoint $x=1$: alternating harmonic series(c) 端点 $x=1$：交错调和级数 M1·A1

At $x=1$ the series becomes $\sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{n}=1-\tfrac{1}{2}+\tfrac{1}{3}-\cdots$ (M1), which converges by the alternating series test: the terms $1/n$ are positive, strictly decreasing, and tend to $0$. (A1)在 $x=1$ 处，级数变为 $\sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{n}=1-\tfrac{1}{2}+\tfrac{1}{3}-\cdots$ (M1)，由交错级数审敛法收敛：各项 $1/n$ 为正，严格递减，趋于 $0$。(A1)

(d) Endpoint $x=-1$: harmonic series diverges; full interval(d) 端点 $x=-1$：调和级数发散；完整收敛区间 A1

At $x=-1$ the series becomes $\sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}(-1)^{n}}{n}=\sum_{n=1}^{\infty}\dfrac{-1}{n}=-\sum_{n=1}^{\infty}\dfrac{1}{n}$, the divergent harmonic series. So $x=-1$ is excluded. (A1)在 $x=-1$ 处，级数变为 $\sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}(-1)^{n}}{n}=\sum_{n=1}^{\infty}\dfrac{-1}{n}=-\sum_{n=1}^{\infty}\dfrac{1}{n}$，即发散的调和级数，故 $x=-1$ 被排除。(A1)

Interval of convergence: $(-1,\,1]$.收敛区间：$(-1,\,1]$。

Insight. Deriving a Maclaurin series from the definition means computing every coefficient $c_{n}=f^{(n)}(0)/n!$ explicitly, not guessing from a pattern. For $\ln(1+x)$ the pattern $f^{(n)}(0)=(-1)^{n-1}(n-1)!$ is clean, but you must exhibit enough derivatives to justify it. The ratio test gives only the open interval; endpoint work is always separate. The Leibniz-type alternating harmonic series at $x=1$ converges to $\ln 2$ (Abel's theorem confirms this), which is a canonical result worth knowing.从定义推导麦克劳林级数意味着须明确计算每个系数 $c_{n}=f^{(n)}(0)/n!$，而非靠猜测规律。对 $\ln(1+x)$，规律 $f^{(n)}(0)=(-1)^{n-1}(n-1)!$ 很整洁，但需展示足够多的导数加以验证。比值审敛法只给出开区间，端点分析须单独进行。$x=1$ 处的莱布尼茨型交错调和级数收敛到 $\ln 2$（阿贝尔定理确认），这是值得熟记的经典结论。
Q5HARDPROOFTaylor remainder (Lagrange form) and error bound泰勒余项（拉格朗日形式）与误差估计[8 marks]

For $f(x)=e^{x}$: (a) write $T_{4}(x)$ and the Lagrange remainder $R_{4}(x)$; (b) show $|e-T_{4}(1)|<\frac{1}{40}$ using $e^{z}<3$ for $0\le z\le 1$; (c) compute $T_{4}(1)$ and verify the bound numerically.对 $f(x)=e^{x}$：(a) 写出 $T_{4}(x)$ 和余项 $R_{4}(x)$ 的拉格朗日形式；(b) 利用 $0\le z\le 1$ 时 $e^{z}<3$，证明 $|e-T_{4}(1)|<\frac{1}{40}$；(c) 计算 $T_{4}(1)$ 并与真实值比较，验证误差估计。

Answers:答案：  (a) $T_{4}(x)=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+\frac{x^{4}}{24}$  ·  (b) $|R_{4}(1)|\le\frac{3}{120}=\frac{1}{40}$  ·  (c) $T_{4}(1)=\frac{65}{24}\approx 2.708$, error $\approx 0.010<0.025$误差 $\approx 0.010<0.025$

(a) Degree-4 polynomial and Lagrange remainder(a) 4次多项式与拉格朗日余项 B1·B1·A1

All derivatives of $e^{x}$ equal $e^{x}$, so $f^{(n)}(0)=1$ for all $n$. (B1) The degree-4 Maclaurin polynomial is$e^{x}$ 的各阶导数均等于 $e^{x}$，故对所有 $n$ 有 $f^{(n)}(0)=1$。(B1) 4次麦克劳林多项式为

$$ T_{4}(x)=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+\frac{x^{4}}{24}. $$

(B1) The Lagrange form of the remainder is(B1) 余项的拉格朗日形式为

$$ R_{4}(x)=\frac{f^{(5)}(z)}{5!}x^{5}=\frac{e^{z}}{120}x^{5} $$

for some $z$ strictly between $0$ and $x$. (A1)其中 $z$ 严格介于 $0$ 与 $x$ 之间。(A1)

(b) Bounding the remainder at $x=1$(b) 在 $x=1$ 处估计余项 M1·A1·R1

Setting $x=1$: $R_{4}(1)=\dfrac{e^{z}}{120}$ for some $z\in(0,1)$. (M1)令 $x=1$：对某个 $z\in(0,1)$，$R_{4}(1)=\dfrac{e^{z}}{120}$。(M1)

Since $e^{z}<3$ for all $z\in[0,1]$ (as $e<3$):由于对所有 $z\in[0,1]$ 有 $e^{z}<3$（因为 $e<3$）：

$$ |e-T_{4}(1)|=|R_{4}(1)|=\frac{e^{z}}{120}<\frac{3}{120}=\frac{1}{40}. $$

(A1) Hence the degree-4 polynomial approximates $e$ to within $1/40=0.025$. (R1)(A1) 故4次多项式对 $e$ 的近似误差不超过 $1/40=0.025$。(R1)

(c) Explicit value and numerical check(c) 明确计算与数值验证 M1·A1

Evaluate $T_{4}(1)$ with a common denominator of $24$: (M1)通分为 $24$ 计算 $T_{4}(1)$：(M1)

$$ T_{4}(1)=\frac{24+24+12+4+1}{24}=\frac{65}{24}\approx 2.70833. $$

The true value is $e\approx 2.71828$, so the actual error is $|e-T_{4}(1)|\approx 0.00995$. Since $0.00995<0.025=1/40$, the bound is confirmed. (A1)真实值为 $e\approx 2.71828$，实际误差为 $|e-T_{4}(1)|\approx 0.00995$。由于 $0.00995<0.025=1/40$，误差估计得到验证。(A1)

Insight. The Lagrange remainder is the rigorous tool for turning a series truncation into a guaranteed accuracy statement. The key steps are: (1) identify $f^{(n+1)}(z)$ and bound it over the relevant interval, (2) plug in the specific $x$ and the bound, (3) simplify to a clean inequality. Here the bound $e^{z}<3$ on $[0,1]$ is loose enough to be trivial but tight enough to be useful. In practice, reusing the standard Maclaurin library and bounding the remainder is much cheaper than differentiating from scratch.拉格朗日余项是将级数截断转化为精度保证的严格工具。关键步骤：(1) 确定 $f^{(n+1)}(z)$ 并在相关区间给出上界，(2) 代入特定 $x$ 值和上界，(3) 化简为简洁的不等式。此处 $[0,1]$ 上的上界 $e^{z}<3$ 足够粗糙以显而易见，又足够紧以实用。实际操作中，复用标准麦克劳林公式库并估计余项，比从头求导的成本低得多。
Q6HARDPROOFbinomial series; radius and interval of convergence二项级数；收敛半径与收敛区间[8 marks]

Using $(1+x)^{k}=\sum_{n=0}^{\infty}\binom{k}{n}x^{n}$ with $k=-\tfrac{1}{2}$: (a) write $(1+x)^{-1/2}$ as a series (four terms and general coefficient); (b) confirm $R=1$ by the ratio test; (c) substitute $x\mapsto -t^{2}$ to expand $\frac{1}{\sqrt{1-t^{2}}}$.取 $(1+x)^{k}=\sum_{n=0}^{\infty}\binom{k}{n}x^{n}$，$k=-\tfrac{1}{2}$：(a) 将 $(1+x)^{-1/2}$ 写成级数（四项及一般系数）；(b) 用比值审敛法验证 $R=1$；(c) 令 $x\mapsto -t^{2}$，展开 $\frac{1}{\sqrt{1-t^{2}}}$。

Answers:答案：  (a) $1-\frac{1}{2}x+\frac{3}{8}x^{2}-\frac{5}{16}x^{3}+\cdots$, general term一般项 $\frac{(-1)^{n}(2n)!}{4^{n}(n!)^{2}}x^{n}$  ·  (b) $R=1$  ·  (c) $\sum_{n=0}^{\infty}\frac{(2n)!}{4^{n}(n!)^{2}}t^{2n}$, $|t|<1$

(a) Generalized binomial coefficients for $k=-\tfrac{1}{2}$(a) $k=-\tfrac{1}{2}$ 时的广义二项系数 M1·A1·A1·A1

Compute the first four generalized binomial coefficients: (M1)计算前四个广义二项系数：(M1)

$\dbinom{-1/2}{0}=1$; $\dbinom{-1/2}{1}=-\tfrac{1}{2}$; $\dbinom{-1/2}{2}=\dfrac{(-\tfrac{1}{2})(-\tfrac{3}{2})}{2!}=\dfrac{3}{8}$; $\dbinom{-1/2}{3}=\dfrac{(-\tfrac{1}{2})(-\tfrac{3}{2})(-\tfrac{5}{2})}{3!}=-\dfrac{5}{16}$. (A1)

So $(1+x)^{-1/2}=1-\dfrac{1}{2}x+\dfrac{3}{8}x^{2}-\dfrac{5}{16}x^{3}+\cdots$ (A1)故 $(1+x)^{-1/2}=1-\dfrac{1}{2}x+\dfrac{3}{8}x^{2}-\dfrac{5}{16}x^{3}+\cdots$ (A1)

The general coefficient satisfies $\dbinom{-1/2}{n}=\dfrac{(-1)^{n}(2n)!}{4^{n}(n!)^{2}}$ (from the identity $\dbinom{-1/2}{n}=\dfrac{(-1)^{n}}{4^{n}}\dbinom{2n}{n}$). (A1)一般系数满足 $\dbinom{-1/2}{n}=\dfrac{(-1)^{n}(2n)!}{4^{n}(n!)^{2}}$（由恒等式 $\dbinom{-1/2}{n}=\dfrac{(-1)^{n}}{4^{n}}\dbinom{2n}{n}$ 得出）。(A1)

(b) Ratio test confirms $R=1$(b) 比值审敛法验证 $R=1$ M1·A1

Let $a_{n}=\dfrac{(2n)!}{4^{n}(n!)^{2}}|x|^{n}$. Then: (M1)令 $a_{n}=\dfrac{(2n)!}{4^{n}(n!)^{2}}|x|^{n}$，则：(M1)

$$ \frac{a_{n+1}}{a_{n}}=\frac{(2n+2)!}{4^{n+1}((n+1)!)^{2}}\cdot\frac{4^{n}(n!)^{2}}{(2n)!}\,|x|=\frac{(2n+1)(2n+2)}{4(n+1)^{2}}\,|x|=\frac{2n+1}{2(n+1)}\,|x|\longrightarrow |x|. $$

Convergence requires $|x|<1$, so $R=1$. (A1)收敛要求 $|x|<1$，故 $R=1$。(A1)

(c) Substitute $x\mapsto -t^{2}$(c) 令 $x\mapsto -t^{2}$ M1·A1

Replace $x$ by $-t^{2}$; the binomial coefficient $(-1)^{n}$ combines with $(-1)^{n}$ from $(-t^{2})^{n}=(-1)^{n}t^{2n}$: (M1)以 $-t^{2}$ 替换 $x$；二项系数中的 $(-1)^{n}$ 与 $(-t^{2})^{n}=(-1)^{n}t^{2n}$ 中的 $(-1)^{n}$ 合并：(M1)

$$ \frac{1}{\sqrt{1-t^{2}}}=\sum_{n=0}^{\infty}\frac{(-1)^{n}(2n)!}{4^{n}(n!)^{2}}(-1)^{n}t^{2n}=\sum_{n=0}^{\infty}\frac{(2n)!}{4^{n}(n!)^{2}}t^{2n}=1+\frac{1}{2}t^{2}+\frac{3}{8}t^{4}+\frac{5}{16}t^{6}+\cdots $$

The restriction $|-t^{2}|<1$ gives $|t|<1$; interval of convergence $(-1,1)$. (A1)限制条件 $|-t^{2}|<1$ 给出 $|t|<1$；收敛区间为 $(-1,1)$。(A1)

Insight. The binomial series is the one Maclaurin series that does not follow from the standard library by differentiation or substitution alone; you must compute the generalized binomial coefficients $\binom{k}{n}$ directly from the definition. Once you have $(1+x)^{k}$, the substitution trick extends it immediately: replacing $x$ with $-t^{2}$ yields $\frac{1}{\sqrt{1-t^{2}}}$, whose integral is $\arcsin t$, another canonical series derivable from this one.二项级数是唯一无法单纯通过微分或换元从标准公式库推导的麦克劳林级数，必须从定义直接计算广义二项系数 $\binom{k}{n}$。一旦得到 $(1+x)^{k}$，换元技巧立即扩展其用途：以 $-t^{2}$ 替换 $x$ 得 $\frac{1}{\sqrt{1-t^{2}}}$，对其积分即得 $\arcsin t$，这是从本结果可导出的另一个经典级数。
PART III  ·  APPLICATIONS AND SYNTHESIS应用与综合Extended problems · 28 marks综合题 · 28分

Worked Solutions详细解答

Q7MEDIUMAPPLIEDseries manipulation: differentiation and multiplication级数运算：微分与乘法[8 marks]

(a) Differentiate the series for $\sin x$ to recover $\cos x$; (b) write $e^{-x^{2}}$ by substitution; (c) write $xe^{-x^{2}}$ and integrate from $0$ to $t$.(a) 对 $\sin x$ 的级数微分，得到 $\cos x$；(b) 通过换元写出 $e^{-x^{2}}$；(c) 写出 $xe^{-x^{2}}$ 并对 $0$ 到 $t$ 积分。

Answers:答案：  (a) $\cos x=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n)!}$  ·  (b) $\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{n!}$  ·  (c) $\sum_{n=0}^{\infty}\frac{(-1)^{n}t^{2n+2}}{2(n+1)!}$

(a) Term-by-term differentiation(a) 逐项微分 M1·A1·A1

Starting from $\sin x=\sum_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+1}}{(2n+1)!}$, differentiate each term: (M1)从 $\sin x=\sum_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+1}}{(2n+1)!}$ 出发，对每项求导：(M1)

$$ \frac{d}{dx}\left[\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}\right]=\frac{(-1)^{n}(2n+1)x^{2n}}{(2n+1)!}=\frac{(-1)^{n}x^{2n}}{(2n)!}. $$

Therefore $\cos x=\sum_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n}}{(2n)!}=1-\dfrac{x^{2}}{2!}+\dfrac{x^{4}}{4!}-\dfrac{x^{6}}{6!}+\cdots$ (A1)故 $\cos x=\sum_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n}}{(2n)!}=1-\dfrac{x^{2}}{2!}+\dfrac{x^{4}}{4!}-\dfrac{x^{6}}{6!}+\cdots$ (A1)

This series converges for all real $x$ (radius of convergence infinite), consistent with $\cos x$ being an entire function. (A1)该级数对所有实数 $x$ 收敛（收敛半径无穷），与 $\cos x$ 为整函数一致。(A1)

(b) Substitute $x\mapsto -x^{2}$ in $e^{x}$(b) 在 $e^{x}$ 中令 $x\mapsto -x^{2}$ M1·A1

From $e^{x}=\sum_{n=0}^{\infty}\dfrac{x^{n}}{n!}$, replace $x$ by $-x^{2}$: (M1)从 $e^{x}=\sum_{n=0}^{\infty}\dfrac{x^{n}}{n!}$ 出发，以 $-x^{2}$ 替换 $x$：(M1)

$$ e^{-x^{2}}=\sum_{n=0}^{\infty}\frac{(-x^{2})^{n}}{n!}=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{n!}=1-x^{2}+\frac{x^{4}}{2!}-\frac{x^{6}}{3!}+\cdots $$

This series converges for all real $x$. (A1)该级数对所有实数 $x$ 收敛。(A1)

(c) Multiply by $x$, then integrate term by term(c) 乘以 $x$，然后逐项积分 M1·A1·A1

Multiply the series in (b) by $x$: (M1)将 (b) 的级数乘以 $x$：(M1)

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

Integrate term by term from $0$ to $t$:从 $0$ 到 $t$ 逐项积分：

$$ \int_{0}^{t}x\,e^{-x^{2}}\,dx=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}\cdot\frac{t^{2n+2}}{2n+2}=\sum_{n=0}^{\infty}\frac{(-1)^{n}\,t^{2n+2}}{2(n+1)!}. $$

(A1) The first few terms are $\dfrac{t^{2}}{2}-\dfrac{t^{4}}{4}+\dfrac{t^{6}}{12}-\dfrac{t^{8}}{48}+\cdots$ (A1)(A1) 前几项为 $\dfrac{t^{2}}{2}-\dfrac{t^{4}}{4}+\dfrac{t^{6}}{12}-\dfrac{t^{8}}{48}+\cdots$ (A1)

Insight. Differentiation and integration of power series are the two most powerful tools in the series toolkit. Differentiating the sine series to get cosine is a concrete check that term-by-term calculus works inside the radius. The Gaussian integral $\int xe^{-x^{2}}dx$ has the elementary antiderivative $-\tfrac{1}{2}e^{-x^{2}}$, and the series in (c) is exactly the Maclaurin expansion of $\tfrac{1}{2}(1-e^{-t^{2}})$. Always exploit the known Maclaurin library rather than differentiating functions repeatedly from scratch.幂级数的微分与积分是级数工具箱中最强大的两种操作。对正弦级数微分得到余弦级数，具体验证了逐项微积分在收敛半径内有效。高斯积分 $\int xe^{-x^{2}}dx$ 有初等原函数 $-\tfrac{1}{2}e^{-x^{2}}$，(c) 中的级数恰好是 $\tfrac{1}{2}(1-e^{-t^{2}})$ 的麦克劳林展开。应充分利用已知的麦克劳林公式库，而非每次从头对函数反复求导。
Q8HARDAPPLIEDapproximating a definite integral with error bound带误差估计的定积分近似[10 marks]

For $I=\int_{0}^{1/2}\frac{\sin x}{x}\,dx$: (a) expand $\frac{\sin x}{x}$ as a power series; (b) integrate term by term for a series for $I$; (c) find the minimum number of terms to approximate $I$ within $10^{-4}$.对 $I=\int_{0}^{1/2}\frac{\sin x}{x}\,dx$：(a) 将 $\frac{\sin x}{x}$ 展开为幂级数；(b) 逐项积分得到 $I$ 的级数；(c) 求近似 $I$ 至 $10^{-4}$ 精度所需的最少项数。

Answers:答案：  (a) $\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n+1)!}$, all $x$对所有 $x$  ·  (b) $\frac{1}{2}-\frac{1}{144}+\frac{1}{19200}-\cdots$  ·  (c) 2 terms; $I\approx\frac{71}{144}\approx 0.49306$2项；$I\approx\frac{71}{144}\approx 0.49306$

(a) Divide the sine series by $x$(a) 将正弦级数除以 $x$ M1·A1·A1

From $\sin x=\sum_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+1}}{(2n+1)!}$, divide every term by $x\ne 0$: (M1)从 $\sin x=\sum_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+1}}{(2n+1)!}$ 出发，每项除以 $x\ne 0$：(M1)

$$ \frac{\sin x}{x}=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n+1)!}=1-\frac{x^{2}}{6}+\frac{x^{4}}{120}-\frac{x^{6}}{5040}+\cdots $$

(A1) The series converges for all $x$ since the sine series does (ratio test gives radius $\infty$), and the convention $(\sin x)/x=1$ at $x=0$ is consistent with the constant term. (A1)(A1) 由于正弦级数对所有 $x$ 收敛（比值审敛法给出无穷收敛半径），该级数亦然，且约定 $x=0$ 时 $(\sin x)/x=1$ 与常数项一致。(A1)

(b) Integrate term by term(b) 逐项积分 M1·A1·A1

Integrating each term from $0$ to $\tfrac{1}{2}$: (M1)对每项从 $0$ 到 $\tfrac{1}{2}$ 积分：(M1)

$$ I=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}\cdot\frac{(1/2)^{2n+1}}{2n+1}=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!\,(2n+1)\,2^{2n+1}}. $$

Computing the first four terms explicitly: (A1)明确计算前四项：(A1)

$n=0$: $\dfrac{1}{1!\cdot 1\cdot 2}=\dfrac{1}{2}$.   $n=1$: $\dfrac{-1}{3!\cdot 3\cdot 8}=\dfrac{-1}{144}$.   $n=2$: $\dfrac{1}{5!\cdot 5\cdot 32}=\dfrac{1}{19200}$.   $n=3$: $\dfrac{-1}{7!\cdot 7\cdot 128}=\dfrac{-1}{4515840}$.

So $I=\dfrac{1}{2}-\dfrac{1}{144}+\dfrac{1}{19200}-\dfrac{1}{4515840}+\cdots$ (A1)故 $I=\dfrac{1}{2}-\dfrac{1}{144}+\dfrac{1}{19200}-\dfrac{1}{4515840}+\cdots$ (A1)

(c) Alternating series estimation theorem(c) 交错级数估计定理 M1·A1·M1·A1

The series for $I$ is alternating with terms $b_{n}=\dfrac{1}{(2n+1)!\,(2n+1)\,2^{2n+1}}$ decreasing to $0$. (M1) By the alternating series estimation theorem, the error when the partial sum stops after $N$ terms (i.e., uses $n=0,1,\ldots,N-1$) is at most $|b_{N}|$.$I$ 的级数为交错级数，各项 $b_{n}=\dfrac{1}{(2n+1)!\,(2n+1)\,2^{2n+1}}$ 递减趋于 $0$。(M1) 由交错级数估计定理，取前 $N$ 项（即 $n=0,1,\ldots,N-1$）的部分和，误差至多为 $|b_{N}|$。

We need $|b_{N}|<10^{-4}$. Check: $|b_{2}|=\dfrac{1}{19200}\approx 5.2\times 10^{-5}<10^{-4}$. (A1)需要 $|b_{N}|<10^{-4}$。检验：$|b_{2}|=\dfrac{1}{19200}\approx 5.2\times 10^{-5}<10^{-4}$。(A1)

And $|b_{1}|=\dfrac{1}{144}\approx 6.9\times 10^{-3}>10^{-4}$, so stopping after $n=0$ is insufficient. (M1)而 $|b_{1}|=\dfrac{1}{144}\approx 6.9\times 10^{-3}>10^{-4}$，故仅取 $n=0$ 项不够。(M1)

Therefore the minimum is $N=2$ terms (the $n=0$ and $n=1$ terms), giving error at most $|b_{2}|\approx 5.2\times 10^{-5}<10^{-4}$:因此最少需 $N=2$ 项（即 $n=0$ 和 $n=1$ 两项），误差至多为 $|b_{2}|\approx 5.2\times 10^{-5}<10^{-4}$：

$$ I\approx \frac{1}{2}-\frac{1}{144}=\frac{72-1}{144}=\frac{71}{144}\approx 0.49306. $$

(A1)

Insight. The alternating series estimation theorem is the engine behind all series-based integral approximations: once you verify the series is alternating with decreasing terms, the first neglected term is an upper bound on the error. Here the $n=2$ term is already below $10^{-4}$, so only two non-zero terms are needed. Recognising that $(\sin x)/x$ is entire (the removable singularity at $0$ is not a real obstruction) is what lets you integrate the series over any finite interval without checking the radius of convergence.交错级数估计定理是所有基于级数的积分近似的核心工具：一旦验证级数为各项递减的交错级数，第一个被舍弃的项即为误差上界。此处 $n=2$ 项已低于 $10^{-4}$，故只需两个非零项。认识到 $(\sin x)/x$ 为整函数（$0$ 处的可去奇点并非实质障碍），使得在任意有限区间上积分该级数无需检验收敛半径。
Q9HARDAPPLIEDseries-based limits; L'Hopital avoidance级数法求极限；避免使用洛必达法则[6 marks]

Using series only: (a) evaluate $\lim_{x\to 0}\frac{e^{x}-1-x}{x^{2}}$; (b) evaluate $\lim_{x\to 0}\frac{\cos x - 1 + \frac{1}{2}x^{2}}{x^{4}}$.仅用级数方法：(a) 计算 $\lim_{x\to 0}\frac{e^{x}-1-x}{x^{2}}$；(b) 计算 $\lim_{x\to 0}\frac{\cos x - 1 + \frac{1}{2}x^{2}}{x^{4}}$。

Answers:答案：  (a) $\tfrac{1}{2}$  ·  (b) $\tfrac{1}{24}$

(a) Expand $e^{x}$ to second order(a) 将 $e^{x}$ 展开至二阶 M1·A1·R1

Write $e^{x}=1+x+\dfrac{x^{2}}{2}+\dfrac{x^{3}}{6}+\cdots$ (M1), so写出 $e^{x}=1+x+\dfrac{x^{2}}{2}+\dfrac{x^{3}}{6}+\cdots$ (M1)，于是

$$ e^{x}-1-x=\frac{x^{2}}{2}+\frac{x^{3}}{6}+\cdots=x^{2}\!\left(\frac{1}{2}+\frac{x}{6}+\cdots\right). $$

Dividing by $x^{2}$ and letting $x\to 0$:除以 $x^{2}$ 并令 $x\to 0$：

$$ \lim_{x\to 0}\frac{e^{x}-1-x}{x^{2}}=\frac{1}{2}+0+\cdots=\frac{1}{2}. $$

(A1) The terms beyond $x^{2}$ vanish as $x\to 0$, confirming the limit equals the coefficient of $x^{2}$ in $e^{x}$. (R1)(A1) $x^{2}$ 以上的项在 $x\to 0$ 时消失，确认极限等于 $e^{x}$ 中 $x^{2}$ 的系数。(R1)

(b) Expand $\cos x$ to fourth order(b) 将 $\cos x$ 展开至四阶 M1·A1·R1

Write $\cos x=1-\dfrac{x^{2}}{2}+\dfrac{x^{4}}{24}-\dfrac{x^{6}}{720}+\cdots$ (M1), so写出 $\cos x=1-\dfrac{x^{2}}{2}+\dfrac{x^{4}}{24}-\dfrac{x^{6}}{720}+\cdots$ (M1)，于是

$$ \cos x - 1 + \frac{x^{2}}{2}=\frac{x^{4}}{24}-\frac{x^{6}}{720}+\cdots=x^{4}\!\left(\frac{1}{24}-\frac{x^{2}}{720}+\cdots\right). $$

Dividing by $x^{4}$ and letting $x\to 0$:除以 $x^{4}$ 并令 $x\to 0$：

$$ \lim_{x\to 0}\frac{\cos x - 1 + \frac{1}{2}x^{2}}{x^{4}}=\frac{1}{24}. $$

(A1) Applying L'Hopital four times to this $0/0$ form would require tracking many derivatives; the series method reads off the answer directly from the fourth-order Taylor coefficient. (R1)(A1) 对这个 $0/0$ 型四次应用洛必达法则需追踪大量导数，而级数方法直接从四阶泰勒系数读出答案。(R1)

Insight. Series limits work by expanding the numerator as a power series in $x$, factoring out the lowest power of $x$ present, then cancelling with the denominator. The limit is simply the lowest-order coefficient that survives the cancellation. For (a) the critical coefficient is $\tfrac{1}{2}$ (the $x^{2}$ coefficient of $e^{x}$); for (b) it is $\tfrac{1}{24}$ (the $x^{4}$ coefficient of $\cos x$). These are exactly the Taylor coefficients $f^{(n)}(0)/n!$, so knowing the Maclaurin library by heart makes these limits instantaneous compared with repeated L'Hopital applications.级数法求极限：将分子展开为 $x$ 的幂级数，提取最低次幂，与分母约分。极限即约分后留存的最低阶系数。(a) 的关键系数为 $\tfrac{1}{2}$（$e^{x}$ 的 $x^{2}$ 系数）；(b) 为 $\tfrac{1}{24}$（$\cos x$ 的 $x^{4}$ 系数）。这些恰好是泰勒系数 $f^{(n)}(0)/n!$，熟记麦克劳林公式库后，这类极限远比反复使用洛必达法则更快捷。
Q10HARDAPPLIEDnumerical approximation with guaranteed accuracy; remainder bound有精度保证的数值近似；余项估计[4 marks]

Approximate $e^{-0.1}$ by $T_{n}(-0.1)$. Using $e^{z}<2$ for $-0.1\le z\le 0$: (a) find the smallest $n$ with $|e^{-0.1}-T_{n}(-0.1)|<5\times 10^{-6}$; (b) state $T_{n}(-0.1)$ to 7 significant figures.用 $T_{n}(-0.1)$ 近似 $e^{-0.1}$，利用 $-0.1\le z\le 0$ 时 $e^{z}<2$：(a) 求满足 $|e^{-0.1}-T_{n}(-0.1)|<5\times 10^{-6}$ 的最小 $n$；(b) 将 $T_{n}(-0.1)$ 精确到7位有效数字。

Answers:答案：  (a) $n=4$  ·  (b) $T_{4}(-0.1)\approx 0.9048375$

(a) Lagrange bound and the minimum $n$(a) 拉格朗日余项估计与最小 $n$ M1·A1·A1

The Lagrange remainder for $e^{x}$ at $x=-0.1$ is $|R_{n}(-0.1)|=\dfrac{e^{z}(0.1)^{n+1}}{(n+1)!}$ for some $z\in(-0.1,0)$. (M1) Using $e^{z}<2$:$e^{x}$ 在 $x=-0.1$ 处的拉格朗日余项为 $|R_{n}(-0.1)|=\dfrac{e^{z}(0.1)^{n+1}}{(n+1)!}$，其中 $z\in(-0.1,0)$。(M1) 利用 $e^{z}<2$：

$$ |R_{n}(-0.1)|<\frac{2\,(0.1)^{n+1}}{(n+1)!}. $$

Test successive values:逐一检验各值：

$n=1$: $\dfrac{2\,(0.01)}{2}=0.01$. Too large.偏大。

$n=2$: $\dfrac{2\,(0.001)}{6}\approx 3.33\times 10^{-4}$. Too large.偏大。

$n=3$: $\dfrac{2\,(0.0001)}{24}\approx 8.33\times 10^{-6}$. Too large.偏大。

$n=4$: $\dfrac{2\,(0.00001)}{120}=\dfrac{2\times 10^{-5}}{120}\approx 1.67\times 10^{-7}<5\times 10^{-6}$. (A1)

The smallest $n$ is $\boxed{4}$. (A1)最小的 $n$ 为 $\boxed{4}$。(A1)

(b) Compute $T_{4}(-0.1)$(b) 计算 $T_{4}(-0.1)$ A1

$T_{4}(-0.1)=1+(-0.1)+\dfrac{(-0.1)^{2}}{2}+\dfrac{(-0.1)^{3}}{6}+\dfrac{(-0.1)^{4}}{24}$

$$ =1-0.1+0.005-\frac{0.001}{6}+\frac{0.0001}{24} $$ $$ =1-0.1+0.005-0.00016\overline{6}+0.000004166\overline{6}\approx 0.9048375. $$

(A1) The true value is $e^{-0.1}\approx 0.9048374$, confirming the error is below $5\times 10^{-6}$.(A1) 真实值为 $e^{-0.1}\approx 0.9048374$，确认误差低于 $5\times 10^{-6}$。

Insight. The Lagrange remainder gives a computable upper bound on the error, so you can certify accuracy without knowing the true value. The strategy is to test $n=1,2,3,\ldots$ until the bound drops below the tolerance. Here the denominator $(n+1)!$ grows much faster than $(0.1)^{n+1}$ shrinks, so convergence is rapid; only $n=4$ is needed for six decimal places. This is the production workflow whenever a numerical answer must meet a stated precision: series expansion, remainder bound, systematic search for the cutoff $n$.拉格朗日余项给出可计算的误差上界，无需知道真实值即可认证精度。策略是依次检验 $n=1,2,3,\ldots$，直到上界低于容许误差。此处分母 $(n+1)!$ 增长远快于 $(0.1)^{n+1}$ 缩小，故收敛迅速；精确到六位小数只需 $n=4$。这是数值答案须满足精度要求时的标准流程：级数展开，余项估计，系统寻找截断阶 $n$。