Unit B7: Power, Taylor, and Maclaurin Series单元 B7：幂级数、泰勒级数与麦克劳林级数

Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{(x-2)^n}{n}$.求 $\displaystyle\sum_{n=1}^{\infty}\frac{(x-2)^n}{n}$ 的收敛区间。

Apply the ratio test to the full term $a_n=(x-2)^n/n$:对整项 $a_n=(x-2)^n/n$ 使用比值判别法：

$$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\left|\frac{(x-2)^{n+1}}{n+1}\cdot\frac{n}{(x-2)^n}\right|=|x-2|\lim_{n\to\infty}\frac{n}{n+1}=|x-2|.$$

Convergence requires $|x-2|<1$, so $R=1$ and the open interval is $1收敛要求 $|x-2|<1$，所以 $R=1$，开区间为 $1

At $x=3$: the series is $\sum 1/n$, the harmonic series, which diverges. At $x=1$: the series is $\sum (-1)^n/n$, which converges by the alternating series test. Hence the interval of convergence is $[1,3)$.在 $x=3$ 处：级数为 $\sum 1/n$，即调和级数，发散。在 $x=1$ 处：级数为 $\sum (-1)^n/n$，由交错级数判别法收敛。因此收敛区间为 $[1,3)$。

Find the radius of convergence of $\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!}$.求 $\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!}$ 的收敛半径。

With $c_n=1/n!$,取 $c_n=1/n!$，

$$\frac{1}{R}=\lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right|=\lim_{n\to\infty}\frac{n!}{(n+1)!}=\lim_{n\to\infty}\frac{1}{n+1}=0.$$

Since $1/R=0$ we have $R=\infty$: the series converges for every real $x$. This series is the exponential function, as Section 5 confirms.由于 $1/R=0$，故 $R=\infty$：级数对每一个实数 $x$ 都收敛。正如第 5 节将确认的，这个级数就是指数函数。

Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n x^{2n}}{4^n\, n}$. Here only even powers appear, so many of the formal coefficients $c_n$ are zero and the formula $1/R=\lim|c_{n+1}/c_n|$ does not apply directly. Apply the ratio test to the whole term instead.求 $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n x^{2n}}{4^n\, n}$ 的收敛区间。这里只出现偶次幂，所以许多形式系数 $c_n$ 为零，公式 $1/R=\lim|c_{n+1}/c_n|$ 不能直接套用。改为对整项使用比值判别法。

Let $a_n=(-1)^n x^{2n}/(4^n n)$. Then令 $a_n=(-1)^n x^{2n}/(4^n n)$。则

$$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\left|\frac{x^{2n+2}}{4^{n+1}(n+1)}\cdot\frac{4^n n}{x^{2n}}\right|=\frac{x^2}{4}\lim_{n\to\infty}\frac{n}{n+1}=\frac{x^2}{4}.$$

Convergence requires $x^2/4<1$, that is $|x|<2$, so $R=2$. At $x=\pm 2$ we have $x^{2n}=4^n$, and the series collapses to $\sum (-1)^n/n$, the alternating harmonic series, which converges. Both endpoints survive, so the interval of convergence is the closed interval $[-2,2]$.收敛要求 $x^2/4<1$，即 $|x|<2$，所以 $R=2$。在 $x=\pm 2$ 处有 $x^{2n}=4^n$，级数退化为 $\sum (-1)^n/n$，即交错调和级数，收敛。两个端点都通过检验，因此收敛区间为闭区间 $[-2,2]$。

The "three cases" theorem rests on one comparison fact. Abel’s lemma: if $\sum c_n (x_0-a)^n$ converges at some point $x_0\neq a$, then $\sum c_n (x-a)^n$ converges absolutely for every $x$ with $|x-a|<|x_0-a|$."三种情形"定理依赖于一个比较事实。阿贝尔引理（Abel’s lemma）：若 $\sum c_n (x_0-a)^n$ 在某点 $x_0\neq a$ 处收敛，则对每一个满足 $|x-a|<|x_0-a|$ 的 $x$，$\sum c_n (x-a)^n$ 都绝对收敛。

Proof. Convergence at $x_0$ forces the terms $c_n (x_0-a)^n\to 0$, so they are bounded: there is $M$ with $|c_n (x_0-a)^n|\le M$ for all $n$. Now fix $x$ with $r:=|x-a|/|x_0-a|<1$. Then证明。在 $x_0$ 处的收敛迫使各项 $c_n (x_0-a)^n\to 0$，于是它们有界：存在 $M$，对所有 $n$ 都有 $|c_n (x_0-a)^n|\le M$。现固定 $x$，令 $r:=|x-a|/|x_0-a|<1$。则

$$|c_n (x-a)^n|=|c_n (x_0-a)^n|\cdot\left|\frac{x-a}{x_0-a}\right|^n\le M\,r^n.$$

Since $\sum M r^n$ is a convergent geometric series ($0\le r<1$), the comparison test gives absolute convergence of $\sum c_n (x-a)^n$. $\blacksquare$由于 $\sum M r^n$ 是收敛的几何级数（$0\le r<1$），由比较判别法得 $\sum c_n (x-a)^n$ 绝对收敛。$\blacksquare$

The lemma says: convergence at one point propagates inward to every closer point. So the set of convergence points has no holes; it is an interval about $a$. Taking $R$ to be the supremum of $|x_0-a|$ over all convergence points $x_0$ produces exactly the three cases, with divergence guaranteed beyond $R$ by the same boundedness argument run in reverse.该引理说：一点处的收敛会向内传播到每一个更近的点。所以收敛点集合没有空洞，它是关于 $a$ 的一个区间。取 $R$ 为所有收敛点 $x_0$ 上 $|x_0-a|$ 的上确界，就恰好得到三种情形，而把同一个有界性论证反向运行，便保证了在 $R$ 之外发散。

Represent $f(x)=\dfrac{1}{1+x^2}$ as a power series and give its radius.把 $f(x)=\dfrac{1}{1+x^2}$ 表示为幂级数，并给出其半径。

Write $1+x^2 = 1-(-x^2)$, so with $u=-x^2$,写成 $1+x^2 = 1-(-x^2)$，于是取 $u=-x^2$，

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

Convergence requires $|-x^2|<1$, that is $|x|<1$, so $R=1$.收敛要求 $|-x^2|<1$，即 $|x|<1$，所以 $R=1$。

Represent $f(x)=\dfrac{1}{3-x}$ as a power series centered at $0$.把 $f(x)=\dfrac{1}{3-x}$ 表示为以 $0$ 为中心的幂级数。

Factor a $3$ out of the denominator so the leading term is $1$:从分母中提取一个 $3$，使首项为 $1$：

$$\frac{1}{3-x}=\frac{1}{3}\cdot\frac{1}{1-\frac{x}{3}}=\frac{1}{3}\sum_{n=0}^{\infty}\left(\frac{x}{3}\right)^n=\sum_{n=0}^{\infty}\frac{x^n}{3^{n+1}}.$$

The condition $\left|\frac{x}{3}\right|<1$ gives $|x|<3$, so $R=3$.条件 $\left|\frac{x}{3}\right|<1$ 给出 $|x|<3$，所以 $R=3$。

Represent $f(x)=\dfrac{x}{x^2-x-2}$ as a power series centered at $0$ and state the radius.把 $f(x)=\dfrac{x}{x^2-x-2}$ 表示为以 $0$ 为中心的幂级数，并给出半径。

Factor the denominator: $x^2-x-2=(x-2)(x+1)$. Decompose into partial fractions, $\dfrac{x}{(x-2)(x+1)}=\dfrac{A}{x-2}+\dfrac{B}{x+1}$. Clearing denominators gives $x=A(x+1)+B(x-2)$. Setting $x=2$ yields $2=3A$, so $A=\tfrac23$; setting $x=-1$ yields $-1=-3B$, so $B=\tfrac13$.分解分母：$x^2-x-2=(x-2)(x+1)$。做部分分式分解 $\dfrac{x}{(x-2)(x+1)}=\dfrac{A}{x-2}+\dfrac{B}{x+1}$。去分母得 $x=A(x+1)+B(x-2)$。令 $x=2$ 得 $2=3A$，故 $A=\tfrac23$；令 $x=-1$ 得 $-1=-3B$，故 $B=\tfrac13$。

Now expand each piece around $0$ in geometric form. For the first, factor out $-2$ so the leading constant is $1$:现在把每一块在 $0$ 附近展成几何级数形式。对第一块，提取 $-2$，使首项常数为 $1$：

$$\frac{2/3}{x-2}=\frac{2}{3}\cdot\frac{1}{x-2}=-\frac{1}{3}\cdot\frac{1}{1-\frac{x}{2}}=-\frac{1}{3}\sum_{n=0}^{\infty}\frac{x^n}{2^n},\qquad |x|<2.$$

For the second piece, $\dfrac{1/3}{x+1}=\dfrac13\cdot\dfrac{1}{1-(-x)}=\dfrac13\sum_{n=0}^{\infty}(-1)^n x^n$, valid for $|x|<1$. Adding the two series term by term,对第二块，$\dfrac{1/3}{x+1}=\dfrac13\cdot\dfrac{1}{1-(-x)}=\dfrac13\sum_{n=0}^{\infty}(-1)^n x^n$，在 $|x|<1$ 时成立。把两个级数逐项相加，

$$\frac{x}{x^2-x-2}=\sum_{n=0}^{\infty}\left(\frac{(-1)^n}{3}-\frac{1}{3\cdot 2^n}\right)x^n.$$

The combined series converges where both pieces do, namely on the smaller interval $|x|<1$, so $R=1$. The radius equals the distance from the center $0$ to the nearest singularity, here $x=-1$.合并后的级数在两块都收敛之处收敛，即在较小的区间 $|x|<1$ 上，所以 $R=1$。半径等于从中心 $0$ 到最近奇点的距离，这里是 $x=-1$。

Could the same function have two different power series about the same center? No. Suppose $\sum c_n (x-a)^n$ and $\sum d_n (x-a)^n$ both converge to $f(x)$ on some interval $|x-a|<\rho$ with $\rho>0$. Their difference $\sum (c_n-d_n)(x-a)^n$ converges to $0$ there.同一个函数能否在同一中心处有两个不同的幂级数？不能。设 $\sum c_n (x-a)^n$ 与 $\sum d_n (x-a)^n$ 在某区间 $|x-a|<\rho$（$\rho>0$）上都收敛到 $f(x)$。它们的差 $\sum (c_n-d_n)(x-a)^n$ 在该区间上收敛到 $0$。

Set $x=a$: every term past the constant vanishes, forcing $c_0-d_0=0$. By Section 3 the zero series may be differentiated term by term; differentiating once and setting $x=a$ forces $c_1-d_1=0$. Differentiating $n$ times and evaluating at $a$ leaves only the $n$th term, giving $n!(c_n-d_n)=0$, hence $c_n=d_n$ for every $n$. So whichever route you use to build a series (substitution, factoring, partial fractions, or the coefficient formula of Section 4), you arrive at the same series. That is why the answer styles in the examples above can be freely interchanged.令 $x=a$：除常数项外每一项都消失，迫使 $c_0-d_0=0$。由第 3 节，这个零级数可以逐项求导；求一次导再令 $x=a$，迫使 $c_1-d_1=0$。求 $n$ 次导并在 $a$ 处取值，只剩第 $n$ 项，得 $n!(c_n-d_n)=0$，故对每个 $n$ 都有 $c_n=d_n$。所以无论你用哪条途径构造级数（代入、提取公因子、部分分式，或第 4 节的系数公式），都会得到同一个级数。这正是上面各例中的不同写法可以自由互换的原因。

Derive the power series for $\arctan x$.推导 $\arctan x$ 的幂级数。

From Example 2.1, $\dfrac{1}{1+t^2}=\sum_{n=0}^{\infty}(-1)^n t^{2n}$ for $|t|<1$. Since $\arctan x=\int_0^x \dfrac{dt}{1+t^2}$, integrate term by term:由例题 2.1，当 $|t|<1$ 时 $\dfrac{1}{1+t^2}=\sum_{n=0}^{\infty}(-1)^n t^{2n}$。因为 $\arctan x=\int_0^x \dfrac{dt}{1+t^2}$，逐项积分：

$$\arctan x=\int_0^x\sum_{n=0}^{\infty}(-1)^n t^{2n}\,dt=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1}=x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots,$$

valid for $|x|<1$. The lower limit $0$ makes the constant of integration zero since $\arctan 0=0$.在 $|x|<1$ 时成立。下限 $0$ 使积分常数为零，因为 $\arctan 0=0$。

Use $\dfrac{1}{1-x}=\sum_{n=0}^{\infty} x^n$ to find a closed form for $\sum_{n=1}^{\infty} n x^{n-1}$.利用 $\dfrac{1}{1-x}=\sum_{n=0}^{\infty} x^n$ 求 $\sum_{n=1}^{\infty} n x^{n-1}$ 的闭式。

Differentiate both sides term by term on $|x|<1$:在 $|x|<1$ 上对两边逐项求导：

$$\frac{d}{dx}\frac{1}{1-x}=\frac{1}{(1-x)^2},\qquad \frac{d}{dx}\sum_{n=0}^{\infty}x^n=\sum_{n=1}^{\infty}n x^{n-1}.$$

Therefore $\displaystyle\sum_{n=1}^{\infty} n x^{n-1}=\frac{1}{(1-x)^2}$ for $|x|<1$. Multiplying by $x$ gives $\sum n x^n = x/(1-x)^2$, a useful closed form.因此当 $|x|<1$ 时 $\displaystyle\sum_{n=1}^{\infty} n x^{n-1}=\frac{1}{(1-x)^2}$。两边乘以 $x$ 得 $\sum n x^n = x/(1-x)^2$，这是一个有用的闭式。

Derive the series for $\ln(1+x)$, then use it to evaluate $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}$.推导 $\ln(1+x)$ 的级数，再用它求 $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}$ 的值。

Start from $\dfrac{1}{1+t}=\sum_{n=0}^{\infty}(-1)^n t^n$ for $|t|<1$ (substitute $-t$ into the geometric series). Since $\ln(1+x)=\int_0^x \dfrac{dt}{1+t}$, integrate term by term:从 $\dfrac{1}{1+t}=\sum_{n=0}^{\infty}(-1)^n t^n$（$|t|<1$，把 $-t$ 代入几何级数）出发。因为 $\ln(1+x)=\int_0^x \dfrac{dt}{1+t}$，逐项积分：

$$\ln(1+x)=\int_0^x\sum_{n=0}^{\infty}(-1)^n t^n\,dt=\sum_{n=0}^{\infty}(-1)^n\frac{x^{n+1}}{n+1}=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots,$$

with constant of integration zero since $\ln 1=0$. The radius is $1$. Now examine the endpoint $x=1$. The numerical series $\sum_{n=1}^{\infty}(-1)^{n-1}/n$ converges by the alternating series test, and Abel’s theorem guarantees the power series is continuous up to a convergent endpoint, so its value there is $\lim_{x\to 1^-}\ln(1+x)=\ln 2$. Therefore积分常数为零，因为 $\ln 1=0$。半径为 $1$。现在考察端点 $x=1$。数项级数 $\sum_{n=1}^{\infty}(-1)^{n-1}/n$ 由交错级数判别法收敛，而阿贝尔定理（Abel’s theorem）保证幂级数在收敛端点处连续，所以它在该处的值为 $\lim_{x\to 1^-}\ln(1+x)=\ln 2$。因此

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

Why does integration leave $R$ unchanged yet possibly widen the interval at an endpoint? Differentiation multiplies the $n$th coefficient by $n$ and integration divides it by $n+1$. Either way the factor behaves like a fixed power of $n$, and $\lim_{n\to\infty} n^{1/n}=1$, so the root-test value $\limsup |c_n|^{1/n}$, hence $R=1/\limsup|c_n|^{1/n}$, is untouched. Endpoints are a different matter, because there $|x-a|=R$ and convergence is decided by the size of the coefficients, not by $R$.为什么积分不改变 $R$，却可能在端点处把区间扩大？求导把第 $n$ 个系数乘以 $n$，积分把它除以 $n+1$。无论哪种，该因子都表现得像 $n$ 的某个固定幂次，而 $\lim_{n\to\infty} n^{1/n}=1$，所以根值判别法的值 $\limsup |c_n|^{1/n}$，从而 $R=1/\limsup|c_n|^{1/n}$，丝毫不变。端点则是另一回事，因为那里 $|x-a|=R$，收敛由系数的大小而非由 $R$ 决定。

Concrete illustration. The geometric series $\sum x^n$ has $R=1$ and diverges at both endpoints. Integrating gives $\sum x^{n+1}/(n+1)$, the series for $-\ln(1-x)$, still $R=1$, but now at $x=-1$ it becomes $\sum (-1)^{n+1}/(n+1)$, which converges. Integration shrank the coefficients by a factor of $1/(n+1)$, enough to rescue one endpoint. Differentiation runs the other way and can lose an endpoint that was previously convergent.具体例证。几何级数 $\sum x^n$ 有 $R=1$，在两个端点都发散。积分得 $\sum x^{n+1}/(n+1)$，即 $-\ln(1-x)$ 的级数，半径仍为 $1$，但现在在 $x=-1$ 处它变成 $\sum (-1)^{n+1}/(n+1)$，收敛。积分把系数缩小了 $1/(n+1)$ 倍，足以救活一个端点。求导走的是相反方向，可能丢掉一个原本收敛的端点。

Find the Taylor series of $f(x)=\ln x$ centered at $a=1$.求 $f(x)=\ln x$ 以 $a=1$ 为中心的泰勒级数。

Compute derivatives: $f(x)=\ln x$, $f'(x)=x^{-1}$, $f''(x)=-x^{-2}$, $f'''(x)=2x^{-3}$, and in general $f^{(n)}(x)=(-1)^{n-1}(n-1)!\,x^{-n}$ for $n\ge 1$. At $a=1$, $f(1)=0$ and $f^{(n)}(1)=(-1)^{n-1}(n-1)!$.计算导数：$f(x)=\ln x$，$f'(x)=x^{-1}$，$f''(x)=-x^{-2}$，$f'''(x)=2x^{-3}$，一般地对 $n\ge 1$ 有 $f^{(n)}(x)=(-1)^{n-1}(n-1)!\,x^{-n}$。在 $a=1$ 处，$f(1)=0$，$f^{(n)}(1)=(-1)^{n-1}(n-1)!$。

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

Find the Maclaurin series of $f(x)=\dfrac{1}{(1-x)^2}$ directly from the coefficient formula, then check it against differentiation of the geometric series.直接用系数公式求 $f(x)=\dfrac{1}{(1-x)^2}$ 的麦克劳林级数，再与对几何级数求导的结果核对。

Compute derivatives at $0$. We have $f(x)=(1-x)^{-2}$, so by the chain rule $f'(x)=2(1-x)^{-3}$, $f''(x)=2\cdot 3(1-x)^{-4}$, and in general $f^{(n)}(x)=(n+1)!\,(1-x)^{-(n+2)}$. At $x=0$ this gives $f^{(n)}(0)=(n+1)!$. The Maclaurin coefficient is therefore在 $0$ 处计算导数。有 $f(x)=(1-x)^{-2}$，故由链式法则 $f'(x)=2(1-x)^{-3}$，$f''(x)=2\cdot 3(1-x)^{-4}$，一般地 $f^{(n)}(x)=(n+1)!\,(1-x)^{-(n+2)}$。在 $x=0$ 处得 $f^{(n)}(0)=(n+1)!$。因此麦克劳林系数为

$$c_n=\frac{f^{(n)}(0)}{n!}=\frac{(n+1)!}{n!}=n+1,$$

so $\dfrac{1}{(1-x)^2}=\sum_{n=0}^{\infty}(n+1)x^n=1+2x+3x^2+4x^3+\cdots$, valid for $|x|<1$. This matches Worked Example 3.2, where the same series arose by differentiating $\sum x^n$ term by term: a reassuring confirmation that the derivative route and the coefficient formula give one and the same series.所以 $\dfrac{1}{(1-x)^2}=\sum_{n=0}^{\infty}(n+1)x^n=1+2x+3x^2+4x^3+\cdots$，在 $|x|<1$ 时成立。这与例题 3.2 一致，那里同一个级数由逐项对 $\sum x^n$ 求导而来：一个令人安心的印证，说明求导路线与系数公式给出的是同一个级数。

Suppose $f(x)=\sum_{k=0}^{\infty} c_k (x-a)^k$ on an interval around $a$. Evaluate at $x=a$: every term with $k\ge 1$ vanishes, so $f(a)=c_0$. Differentiate term by term (legitimate inside the radius, by Section 3):设在 $a$ 附近的某区间上 $f(x)=\sum_{k=0}^{\infty} c_k (x-a)^k$。在 $x=a$ 处取值：所有 $k\ge 1$ 的项都消失，故 $f(a)=c_0$。逐项求导（由第 3 节，在半径内合法）：

$$f'(x)=\sum_{k=1}^{\infty} k\,c_k (x-a)^{k-1},$$

and setting $x=a$ kills all but the $k=1$ term, giving $f'(a)=c_1$. Differentiating $n$ times brings down the factor $k(k-1)\cdots(k-n+1)$, and at $x=a$ only the $k=n$ term survives:令 $x=a$ 后除 $k=1$ 项外全部消去，得 $f'(a)=c_1$。求 $n$ 次导会带下因子 $k(k-1)\cdots(k-n+1)$，而在 $x=a$ 处只剩 $k=n$ 项：

$$f^{(n)}(a)=n(n-1)\cdots 1\cdot c_n=n!\,c_n.$$

Hence $c_n=f^{(n)}(a)/n!$. The representation, if it exists, is unique and equals the Taylor series.于是 $c_n=f^{(n)}(a)/n!$。这个表示若存在，就是唯一的，且等于泰勒级数。

Find the Maclaurin series for $e^{-x^2}$.求 $e^{-x^2}$ 的麦克劳林级数。

Substitute $-x^2$ for $x$ in the series for $e^x$:在 $e^x$ 的级数中把 $x$ 换成 $-x^2$：

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

valid for all $x$. This function has no elementary antiderivative, yet its series integrates term by term effortlessly, as Section 7 uses.对所有 $x$ 成立。这个函数没有初等原函数，但它的级数可以毫不费力地逐项积分，第 7 节会用到这一点。

Expand $\sqrt{1+x}=(1+x)^{1/2}$ through the cubic term.把 $\sqrt{1+x}=(1+x)^{1/2}$ 展开到三次项。

With $k=\tfrac12$, compute the binomial coefficients: $\binom{1/2}{0}=1$, $\binom{1/2}{1}=\tfrac12$, $\binom{1/2}{2}=\tfrac{(1/2)(-1/2)}{2}=-\tfrac18$, $\binom{1/2}{3}=\tfrac{(1/2)(-1/2)(-3/2)}{6}=\tfrac{1}{16}$. Hence取 $k=\tfrac12$，计算二项系数：$\binom{1/2}{0}=1$，$\binom{1/2}{1}=\tfrac12$，$\binom{1/2}{2}=\tfrac{(1/2)(-1/2)}{2}=-\tfrac18$，$\binom{1/2}{3}=\tfrac{(1/2)(-1/2)(-3/2)}{6}=\tfrac{1}{16}$。于是

$$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\cdots,\qquad |x|<1.$$

Find the Maclaurin series of $f(x)=e^x\cos x$ through the $x^4$ term.求 $f(x)=e^x\cos x$ 的麦克劳林级数，展开到 $x^4$ 项。

Multiply the two known series and collect like powers. Using $e^x=1+x+\tfrac{x^2}{2}+\tfrac{x^3}{6}+\tfrac{x^4}{24}+\cdots$ and $\cos x=1-\tfrac{x^2}{2}+\tfrac{x^4}{24}-\cdots$, the coefficient of $x^m$ in the product is the convolution $\sum_{j=0}^{m} a_j b_{m-j}$ of the two coefficient sequences. Term by term:把两个已知级数相乘并合并同次幂。用 $e^x=1+x+\tfrac{x^2}{2}+\tfrac{x^3}{6}+\tfrac{x^4}{24}+\cdots$ 与 $\cos x=1-\tfrac{x^2}{2}+\tfrac{x^4}{24}-\cdots$，乘积中 $x^m$ 的系数是两个系数列的卷积 $\sum_{j=0}^{m} a_j b_{m-j}$。逐项计算：

$$[x^0]:\;1,\quad [x^1]:\;1,\quad [x^2]:\;\tfrac12-\tfrac12=0,\quad [x^3]:\;\tfrac16-\tfrac12=-\tfrac13,\quad [x^4]:\;\tfrac{1}{24}-\tfrac14+\tfrac{1}{24}=-\tfrac16.$$

Hence $e^x\cos x=1+x-\dfrac{x^3}{3}-\dfrac{x^4}{6}+\cdots$. Both factors converge for all $x$, so the product does too.于是 $e^x\cos x=1+x-\dfrac{x^3}{3}-\dfrac{x^4}{6}+\cdots$。两个因子对所有 $x$ 都收敛，所以乘积也收敛。

For $f(x)=\cos x$ the derivatives cycle: $f'(x)=-\sin x$, $f''(x)=-\cos x$, $f'''(x)=\sin x$, $f^{(4)}(x)=\cos x$, period four. At $x=0$ this gives $f(0)=1$, $f'(0)=0$, $f''(0)=-1$, $f'''(0)=0$, and so on, so the odd coefficients vanish and the even ones alternate in sign:对 $f(x)=\cos x$，导数循环出现：$f'(x)=-\sin x$，$f''(x)=-\cos x$，$f'''(x)=\sin x$，$f^{(4)}(x)=\cos x$，周期为四。在 $x=0$ 处得 $f(0)=1$，$f'(0)=0$，$f''(0)=-1$，$f'''(0)=0$，依此类推，于是奇数项系数消失，偶数项系数符号交替：

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

To prove this equals $\cos x$ rather than merely being its formal series, bound the Lagrange remainder. Every derivative of cosine is $\pm\sin$ or $\pm\cos$, so $|f^{(n+1)}(c)|\le 1$ for all $c$ and all $n$. Hence要证明这等于 $\cos x$，而不只是它的形式级数，需对拉格朗日余项作界估计。余弦的每一阶导数都是 $\pm\sin$ 或 $\pm\cos$，所以对所有 $c$ 与所有 $n$ 有 $|f^{(n+1)}(c)|\le 1$。于是

$$|R_n(x)|=\frac{|f^{(n+1)}(c)|}{(n+1)!}|x|^{n+1}\le\frac{|x|^{n+1}}{(n+1)!}\xrightarrow[n\to\infty]{}0$$

for every fixed $x$, because the factorial outgrows any fixed power. Therefore $R_n(x)\to 0$ and the series equals $\cos x$ on all of $\mathbb{R}$. The identical argument, with the same uniform bound $1$, validates the sine series everywhere.对每一个固定的 $x$ 都成立，因为阶乘的增长压过任何固定幂次。因此 $R_n(x)\to 0$，级数在整个 $\mathbb{R}$ 上等于 $\cos x$。完全相同的论证，用同一个一致上界 $1$，也在处处验证了正弦级数。

Estimate the error in approximating $e^{0.1}$ by its degree $2$ Taylor polynomial $T_2(x)=1+x+\tfrac{x^2}{2}$ at $a=0$.估计用 $a=0$ 处的 $2$ 次泰勒多项式 $T_2(x)=1+x+\tfrac{x^2}{2}$ 逼近 $e^{0.1}$ 的误差。

For $f(x)=e^x$ all derivatives equal $e^x$. On $[0,0.1]$ we have $f'''(c)=e^c\le e^{0.1}<1.2$. The Lagrange remainder with $n=2$ is对 $f(x)=e^x$，所有导数都等于 $e^x$。在 $[0,0.1]$ 上有 $f'''(c)=e^c\le e^{0.1}<1.2$。取 $n=2$ 的拉格朗日余项为

$$|R_2(0.1)|=\frac{e^c}{3!}(0.1)^3\le \frac{1.2}{6}(0.001)=2\times 10^{-4}.$$

So $T_2(0.1)=1.105$ approximates $e^{0.1}\approx 1.10517$ to within $0.0002$, consistent with the bound.所以 $T_2(0.1)=1.105$ 逼近 $e^{0.1}\approx 1.10517$，误差在 $0.0002$ 以内，与界一致。

How many terms of the Maclaurin series for $\sin x$ are needed to compute $\sin(0.5)$ with error below $10^{-6}$?要把 $\sin(0.5)$ 算到误差小于 $10^{-6}$，需要 $\sin x$ 的麦克劳林级数的多少项？

Two valid bounds are available, and it is instructive to use both. Lagrange bound: every derivative of sine is bounded by $1$, so the degree $n$ remainder satisfies $|R_n(0.5)|\le \dfrac{(0.5)^{n+1}}{(n+1)!}$. Alternating-series bound: the series $\sin x=x-\tfrac{x^3}{3!}+\tfrac{x^5}{5!}-\cdots$ alternates with decreasing terms at $x=0.5$, so the error after any partial sum is at most the first omitted term.有两个有效的界都可用，把两个都用一遍很有启发性。拉格朗日界：正弦的每一阶导数都不超过 $1$，所以 $n$ 次余项满足 $|R_n(0.5)|\le \dfrac{(0.5)^{n+1}}{(n+1)!}$。交错级数界：级数 $\sin x=x-\tfrac{x^3}{3!}+\tfrac{x^5}{5!}-\cdots$ 在 $x=0.5$ 处交错且各项递减，所以任一部分和之后的误差至多等于第一个被略去的项。

Test successive omitted terms at $x=0.5$:在 $x=0.5$ 处逐个检验被略去的项：

$$\frac{(0.5)^5}{5!}=\frac{0.03125}{120}\approx 2.6\times10^{-4},\qquad \frac{(0.5)^7}{7!}=\frac{0.0078125}{5040}\approx 1.55\times10^{-6},\qquad \frac{(0.5)^9}{9!}\approx 4.3\times10^{-9}.$$

The first omitted term below $10^{-6}$ is the $x^9$ term, so keeping through the $x^7$ term, that is $\sin(0.5)\approx 0.5-\tfrac{(0.5)^3}{6}+\tfrac{(0.5)^5}{120}-\tfrac{(0.5)^7}{5040}$, guarantees the target. This gives $0.4794255$, against the true value $0.4794255386\ldots$, well within $10^{-6}$.第一个小于 $10^{-6}$ 的被略去项是 $x^9$ 项，所以保留到 $x^7$ 项，即 $\sin(0.5)\approx 0.5-\tfrac{(0.5)^3}{6}+\tfrac{(0.5)^5}{120}-\tfrac{(0.5)^7}{5040}$，便可保证达到目标。这给出 $0.4794255$，与真值 $0.4794255386\ldots$ 相比，远在 $10^{-6}$ 之内。

Fix any real $x$. For $f(t)=e^t$, every derivative is $e^t$, so on the interval between $0$ and $x$ we have $|f^{(n+1)}(c)|=e^c\le e^{|x|}=:M$, a constant independent of $n$. The Lagrange remainder at $a=0$ then satisfies固定任意实数 $x$。对 $f(t)=e^t$，每一阶导数都是 $e^t$，所以在 $0$ 与 $x$ 之间的区间上有 $|f^{(n+1)}(c)|=e^c\le e^{|x|}=:M$，这是一个与 $n$ 无关的常数。于是 $a=0$ 处的拉格朗日余项满足

$$|R_n(x)|=\frac{e^c}{(n+1)!}|x|^{n+1}\le \frac{M\,|x|^{n+1}}{(n+1)!}.$$

For fixed $x$, the sequence $|x|^{n+1}/(n+1)!$ tends to $0$ as $n\to\infty$ (the factorial dominates any fixed power). Therefore $R_n(x)\to 0$, and $e^x=\sum_{n=0}^{\infty} x^n/n!$ for every real $x$. The same uniform-bound argument, using $|f^{(n+1)}|\le 1$, proves the sine and cosine series everywhere.对固定的 $x$，序列 $|x|^{n+1}/(n+1)!$ 当 $n\to\infty$ 时趋于 $0$（阶乘压过任何固定幂次）。因此 $R_n(x)\to 0$，且对每一个实数 $x$ 都有 $e^x=\sum_{n=0}^{\infty} x^n/n!$。同样的一致界论证，用 $|f^{(n+1)}|\le 1$，便在处处证明了正弦与余弦级数。

Evaluate $\displaystyle\lim_{x\to 0}\frac{\sin x-x}{x^3}$.求 $\displaystyle\lim_{x\to 0}\frac{\sin x-x}{x^3}$。

Use $\sin x=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\cdots$. Then $\sin x-x=-\dfrac{x^3}{6}+\dfrac{x^5}{120}-\cdots$, so用 $\sin x=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\cdots$。则 $\sin x-x=-\dfrac{x^3}{6}+\dfrac{x^5}{120}-\cdots$，所以

$$\frac{\sin x-x}{x^3}=-\frac{1}{6}+\frac{x^2}{120}-\cdots\;\xrightarrow[x\to 0]{}\;-\frac{1}{6}.$$

The series exposes the cubic cancellation immediately, where three applications of L’Hopital’s rule would be required otherwise.级数立刻揭示了三次项的相消，否则就需要连用三次洛必达法则。

Approximate $\displaystyle\int_0^1 \frac{\sin x}{x}\,dx$ to three decimal places.把 $\displaystyle\int_0^1 \frac{\sin x}{x}\,dx$ 逼近到三位小数。

From $\sin x=\sum (-1)^n x^{2n+1}/(2n+1)!$ we get $\dfrac{\sin x}{x}=\sum (-1)^n x^{2n}/(2n+1)!$. Integrating term by term from $0$ to $1$,由 $\sin x=\sum (-1)^n x^{2n+1}/(2n+1)!$ 得 $\dfrac{\sin x}{x}=\sum (-1)^n x^{2n}/(2n+1)!$。从 $0$ 到 $1$ 逐项积分，

$$\int_0^1\frac{\sin x}{x}\,dx=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!\,(2n+1)}=1-\frac{1}{18}+\frac{1}{600}-\frac{1}{35280}+\cdots.$$

This alternates, so the truncation error is at most the first omitted term. Summing through $1/600$ gives $0.94611$, and the next term $1/35280\approx 0.0000283$ confirms the value is $0.946$ to three places.这是交错级数，所以截断误差至多等于第一个被略去的项。求和到 $1/600$ 得 $0.94611$，而下一项 $1/35280\approx 0.0000283$ 证实该值精确到三位小数为 $0.946$。

Solve the initial value problem $y'=y$, $y(0)=1$, by assuming a power series solution.假设有幂级数解，求解初值问题 $y'=y$，$y(0)=1$。

Write $y=\sum_{n=0}^{\infty} a_n x^n$. Then $y'=\sum_{n=1}^{\infty} n a_n x^{n-1}=\sum_{n=0}^{\infty}(n+1)a_{n+1}x^n$ after reindexing. The equation $y'=y$ forces the coefficient of $x^n$ on each side to agree:写 $y=\sum_{n=0}^{\infty} a_n x^n$。则 $y'=\sum_{n=1}^{\infty} n a_n x^{n-1}=\sum_{n=0}^{\infty}(n+1)a_{n+1}x^n$（重新标号后）。方程 $y'=y$ 迫使两边 $x^n$ 的系数相等：

$$(n+1)a_{n+1}=a_n\quad\Longrightarrow\quad a_{n+1}=\frac{a_n}{n+1}.$$

The initial condition gives $a_0=y(0)=1$. Iterating the recursion: $a_1=1$, $a_2=\tfrac12$, $a_3=\tfrac{1}{6}$, and in general $a_n=\tfrac{1}{n!}$. Therefore初始条件给出 $a_0=y(0)=1$。迭代递推：$a_1=1$，$a_2=\tfrac12$，$a_3=\tfrac{1}{6}$，一般地 $a_n=\tfrac{1}{n!}$。因此

$$y=\sum_{n=0}^{\infty}\frac{x^n}{n!}=e^x,$$

recovering the known solution. The power series method, matching coefficients through a recursion, is the gateway to series solutions of equations with no elementary closed form, the central technique of a later differential equations course.恢复了已知的解。幂级数方法通过递推匹配系数，是求那些没有初等闭式的方程的级数解的入口，也是后续微分方程课程的核心技巧。

The four applications above share a structure: an analytic operation on a function (limit, integral, root-finding, differential equation) becomes an algebraic operation on coefficients (cancellation, term by term integration, recursion). One historical payoff is the computation of $\pi$. From Worked Example 3.1, $\arctan x=x-\tfrac{x^3}{3}+\tfrac{x^5}{5}-\cdots$ for $|x|\le 1$. Setting $x=1$ gives the Leibniz formula上面四个应用共享同一结构：对函数的一种分析运算（极限、积分、求根、微分方程）变成对系数的一种代数运算（相消、逐项积分、递推）。一个历史性的回报是 $\pi$ 的计算。由例题 3.1，当 $|x|\le 1$ 时 $\arctan x=x-\tfrac{x^3}{3}+\tfrac{x^5}{5}-\cdots$。令 $x=1$ 得莱布尼茨公式

$$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots.$$

This converges, but agonizingly slowly: the error after $N$ terms is about $1/(2N)$, so four-decimal accuracy needs thousands of terms. The fix is to evaluate $\arctan$ at a small argument, where the series races to convergence. Machin’s identity $\tfrac{\pi}{4}=4\arctan\tfrac15-\arctan\tfrac{1}{239}$ feeds the series tiny inputs, and a handful of terms then pins $\pi$ to many digits. The lesson generalizes: a series is a tool, and choosing where to evaluate it is half the craft.它收敛，却慢得令人发指：$N$ 项之后的误差约为 $1/(2N)$，所以四位小数的精度需要上千项。补救办法是在一个小的自变量处求 $\arctan$，那里级数飞速收敛。马青恒等式 $\tfrac{\pi}{4}=4\arctan\tfrac15-\arctan\tfrac{1}{239}$ 给级数喂进极小的输入，于是寥寥几项就把 $\pi$ 钉到许多位。这一教训可以推广：级数是工具，而选择在何处求值，乃是手艺的一半。