这一讲的两条线

上一讲末尾给了弧长公式，但只是一个定义。这一讲先把它讲清楚：为什么弧长等于 $\int\sqrt{f'^2+g'^2}\,ds$，并用单位圆算出一个经典结论——$\arccos$ 的积分表示。

随后换到另一个话题：函数列的收敛。数列收敛只有一个概念，函数列却有两个：

• 逐点收敛：每个固定的 $x$，数列 $f_n(x)$ 收敛。
• 一致收敛：同一个 $N$ 对所有 $x$ 都管用。

两者差别看似技术性，实则决定了能否交换极限与积分、求导。这一讲给出判别一致收敛的主要工具：Weierstrass M 判别法。

时间轴

弧长：定义与理由

设平面曲线

$$\gamma(s)=\big(f(s),g(s)\big), \qquad s\in[a,b],$$

其中 $f,g$ 可导且导数连续。弧长定义为

$$L=\int_a^b\sqrt{f'(s)^2+g'(s)^2}\;ds .$$

为什么合理。 取 $[a,b]$ 的分割 $a=s_0<s_1<\cdots<s_n=b$。曲线被分成若干小段，第 $i$ 段从 $\gamma(s_{i-1})$ 到 $\gamma(s_i)$。用弦长近似这段弧长：

$$\big|\gamma(s_i)-\gamma(s_{i-1})\big| =\sqrt{\big(f(s_i)-f(s_{i-1})\big)^2+\big(g(s_i)-g(s_{i-1})\big)^2}.$$

对 $f$ 与 $g$ 分别用中值定理：存在 $\xi_i,\eta_i\in(s_{i-1},s_i)$，使

$$f(s_i)-f(s_{i-1})=f'(\xi_i)\Delta s_i, \qquad g(s_i)-g(s_{i-1})=g'(\eta_i)\Delta s_i,$$

其中 $\Delta s_i=s_i-s_{i-1}$。代入：

$$\big|\gamma(s_i)-\gamma(s_{i-1})\big| =\sqrt{f'(\xi_i)^2+g'(\eta_i)^2}\;\Delta s_i .$$

把各段相加，得到一个 Riemann 和。由于 $f',g'$ 在紧区间上连续，因而一致连续，当分割足够细时，$\xi_i,\eta_i$ 都落在同一小段内，$\sqrt{f'(\xi_i)^2+g'(\eta_i)^2}$ 与 $\sqrt{f'(s)^2+g'(s)^2}$ 在小段上的差别可以一致地压小。于是 Riemann 和收敛到积分

$$\int_a^b\sqrt{f'(s)^2+g'(s)^2}\;ds .$$

这就是弧长公式的来源：把曲线切成小段，用弦代替弧，再取极限。严格论证要用的正是一致连续性——这与上一讲证明连续函数可积是同一套思路。

例子一：抛物线弧

取

$$\gamma(s)=(s,s^2), \qquad s\in[0,1].$$

则 $f'=1$、$g'=2s$，所以

$$L=\int_0^1\sqrt{1+4s^2}\;ds .$$

这个积分可以用换元 $2s=\sinh t$ 或 $2s=\tan\theta$ 算出，课程只要求写出积分式。要点是：弧长公式把几何问题变成了一个确定的定积分。

例子二：单位圆与 arccos

单位圆 $x^2+y^2=1$ 的上半部分可以参数化为

$$\gamma(s)=\big(s,\sqrt{1-s^2}\big), \qquad s\in[-1,1].$$

这里 $f(s)=s$，$g(s)=\sqrt{1-s^2}$。求导：

$$f'(s)=1, \qquad g'(s)=\frac{1}{2\sqrt{1-s^2}}\cdot(-2s)=-\frac{s}{\sqrt{1-s^2}} .$$

于是

$$f'(s)^2+g'(s)^2 =1+\frac{s^2}{1-s^2} =\frac{1-s^2+s^2}{1-s^2} =\frac{1}{1-s^2} .$$

因此这段圆的弧长微元是 $\dfrac{ds}{\sqrt{1-s^2}}$。

现在取圆上横坐标为 $x$ 的点，沿圆走到 $(1,0)$，这段弧长记为 $\theta$。因为 $s$ 是横坐标，从 $s=x$ 到 $s=1$：

$$\theta=\int_x^1\frac{ds}{\sqrt{1-s^2}} .$$

几何上，$\theta$ 正是这段弧对应的圆心角，而横坐标 $x=\cos\theta$。所以 $\theta=\arccos x$，得到

$$\boxed{\ \arccos x=\int_x^1\frac{ds}{\sqrt{1-s^2}}\ }$$

这个公式不是背出来的，而是「角度 = 弧长」这一定义加上弧长公式的结果。

顺带求出 arccos 的导数

上式右边的积分下限是 $x$，不是上限。用区间可加改写：

$$\int_x^1\frac{ds}{\sqrt{1-s^2}} = \int_0^1\frac{ds}{\sqrt{1-s^2}} - \int_0^x\frac{ds}{\sqrt{1-s^2}} .$$

第一项是常数，求导为零；第二项由基本定理（第一版）求导得被积函数在 $x$ 处的值。所以

$$\frac{d}{dx}\arccos x =0-\frac{1}{\sqrt{1-x^2}} =-\frac{1}{\sqrt{1-x^2}} .$$

课程特意强调符号：变量在下限时要多一个负号。这来自 $\int_x^1=-\int_1^x$。

逐点收敛

设 $I$ 是一个区间，$(f_n)$ 是 $I$ 上的函数列，$f$ 也是 $I$ 上的函数。称 $f_n\to f$ 逐点收敛，若对每个固定的 $x\in I$，

$$\lim_{n\to\infty}f_n(x)=f(x).$$

展开成 $\varepsilon$ 语言：对每个 $x\in I$ 和每个 $\varepsilon>0$，存在 $N$，使得 $n>N$ 时 $|f_n(x)-f(x)|<\varepsilon$。注意 $N$ 可以依赖 $x$。

例子一。 在 $[0,1]$ 上取 $f_n(x)=x^n$，并令

$$f(x)= \begin{cases} 0, & 0\le x<1,\\ 1, & x=1 . \end{cases}$$

固定 $x<1$，则 $x^n\to0$；固定 $x=1$，则 $1^n=1$。所以 $f_n\to f$ 逐点收敛。每个 $f_n$ 都连续，极限 $f$ 却不连续——逐点收敛不保持连续性。

例子二。 在 $\mathbb R$ 上取

$$f_n(x)=\sum_{k=0}^{n}\frac{x^k}{k!} .$$

对每个固定的 $x$，这是指数级数的部分和，收敛到 $e^x$。所以 $f_n\to e^x$ 逐点收敛。

一致收敛

称 $f_n\to f$ 一致收敛，若对每个 $\varepsilon>0$，存在 $N$，使得

$$n>N \implies |f_n(x)-f(x)|<\varepsilon \quad\text{对一切 }x\in I .$$

区别只有一个：同一个 $N$ 必须对所有 $x$ 有效。

引理。 一致收敛蕴含逐点收敛。

证明。 若一致收敛，给定 $x$ 与 $\varepsilon$，取一致收敛定义中的 $N$；它对所有 $x$ 都有效，当然对这个 $x$ 有效，于是 $n>N$ 时 $|f_n(x)-f(x)|<\varepsilon$。这正是逐点收敛的定义（只是这里的 $N$ 恰好不依赖 $x$）。$\square$

反过来不成立，下面的反例就是证据。

反例：$x^n$ 在 $[0,1]$ 上不一致收敛

图：逐点与一致的量词差异，以及 xⁿ 为什么进不了 ε 带。

已经知道 $f_n(x)=x^n$ 逐点收敛到上面的 $f$（在 $x=1$ 处为 $1$，其余为 $0$）。但不一致收敛。

证明。 每个 $f_n$ 连续，且 $f_n(0)=0$、$f_n(1)=1$。由介值定理，存在 $x_n\in(0,1)$，使

$$f_n(x_n)=x_n^n=\frac12 .$$

（具体地 $x_n=(1/2)^{1/n}$，当 $n\to\infty$ 时 $x_n\to1$。）

而 $f(x_n)=0$，因为 $x_n<1$。于是

$$|f_n(x_n)-f(x_n)|=\frac12 .$$

取 $\varepsilon=\frac14$。若一致收敛，则存在 $N$，使 $n>N$ 时对所有 $x$ 都有 $|f_n(x)-f(x)|<\frac14$。但在 $x=x_n$ 处差值是 $\frac12>\frac14$，矛盾。所以不一致收敛。$\square$

几何上：每个 $f_n$ 都从 $0$ 爬到 $1$，中间必经过 $\frac12$；这个「半高点」随 $n$ 增大不断向右移动到 $1$。极限函数在这些点取 $0$，所以误差始终是 $\frac12$，无论 $n$ 多大。逐点收敛只保证每个固定点的误差最终变小，不保证误差在整段区间上同时变小。

这也解释了为什么 $x^n$ 的极限不连续：一致性正是连续性得以继承的条件。

Weierstrass M 判别法

定理。 设 $(f_n)$ 是区间 $I$ 上的函数列，且存在非负实数 $M_n$，使

$$|f_n(x)|\le M_n \quad\text{对一切 }x\in I,$$

并设级数 $\sum_{n=0}^\infty M_n$ 收敛。则函数项级数 $\sum_{n=0}^\infty f_n(x)$ 在 $I$ 上一致收敛。

证明。 记部分和

$$S_n(x)=\sum_{k=0}^{n}f_k(x).$$

对 $n>m$，

$$S_n(x)-S_m(x)=\sum_{k=m+1}^{n}f_k(x),$$

取绝对值并用三角不等式与假设：

$$|S_n(x)-S_m(x)| \le\sum_{k=m+1}^{n}|f_k(x)| \le\sum_{k=m+1}^{n}M_k \le\sum_{k=m+1}^{\infty}M_k .$$

因为 $\sum M_k$ 收敛，给定 $\varepsilon>0$，存在 $N$，使 $m\ge N$ 时尾部

$$\sum_{k=m+1}^{\infty}M_k<\varepsilon .$$

第一步：逐点收敛。 对每个固定的 $x$，上式说明 $(S_n(x))$ 是 Cauchy 数列，因此收敛；把极限记为 $S(x)$，得到函数 $S$。

第二步：升级为一致收敛。 在不等式

$$|S_n(x)-S_m(x)|\le\sum_{k=m+1}^{\infty}M_k$$

中令 $n\to\infty$（左边趋于 $|S(x)-S_m(x)|$），得

$$|S(x)-S_m(x)|\le\sum_{k=m+1}^{\infty}M_k<\varepsilon \qquad(m\ge N).$$

右边与 $x$ 无关，且对所有 $x$ 成立。这正是 $S_m\to S$ 一致收敛。$\square$

关键是最后一步：尾部的界不含 $x$。这就是「M」的用途——用一列与 $x$ 无关的数的尾部，控制一整族函数的尾部。

应用：指数级数

在 $[-L,L]$ 上考察

$$f_n(x)=\frac{x^n}{n!} .$$

取

$$M_n=\frac{L^n}{n!} .$$

则对 $|x|\le L$，

$$|f_n(x)|=\frac{|x|^n}{n!}\le\frac{L^n}{n!}=M_n .$$

而 $\sum L^n/n!$ 收敛（比值判别法：相邻项之比为 $L/(n+1)\to0$）。由 M 判别法，

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

在 $[-L,L]$ 上一致收敛到 $e^x$。因为 $L$ 任意，它在每个有界区间上一致收敛——但不是在整个 $\mathbb R$ 上一致收敛，原因稍后说明。

幂级数与内闭一致收敛

设

$$\sum_{n=0}^{\infty}a_nx^n$$

是幂级数。定义

$$M=\limsup_{n\to\infty}|a_n|^{1/n}, \qquad R=\frac1M,$$

并约定 $M=0$ 时 $R=\infty$、$M=\infty$ 时 $R=0$。$R$ 称为收敛半径。

已知的事实：

• $|x|<R$ 时级数绝对收敛；
• $|x|>R$ 时级数发散；
• $|x|=R$ 的边界点要单独判断，没有统一结论。

定理。 若 $0<L<R$，则 $\sum a_nx^n$ 在 $[-L,L]$ 上一致收敛。

证明。 对 $|x|\le L$，

$$|a_nx^n|\le|a_n|L^n .$$

需要说明 $\sum|a_n|L^n$ 收敛。由 $M=\limsup|a_n|^{1/n}=1/R$，

$$\limsup_{n\to\infty}\big(|a_n|L^n\big)^{1/n} =ML=\frac{L}{R}<1 .$$

由根值判别法，$\sum|a_n|L^n$ 收敛。取 $M_n=|a_n|L^n$，M 判别法给出在 $[-L,L]$ 上一致收敛。$\square$

注意定理要求 $L$ 严格小于 $R$。在端点 $\pm R$ 处，收敛可能发生也可能不发生，一致性也会失效——例如 $\sum x^n$ 的收敛半径是 $1$，在 $[-L,L]$（$L<1$）上一致收敛，但在 $[0,1)$ 上不一致收敛（与 $x^n$ 的反例同源）。

课程提到，一致收敛之所以重要，是因为它允许逐项积分与逐项求导：若 $f_n\to f$ 一致收敛，则

$$\lim_{n\to\infty}\int_a^b f_n=\int_a^b f,$$

在适当条件下还有 $\lim f_n'=f'$。这是下一讲的主题。

本节结论

• 弧长 $=\int_a^b\sqrt{f'^2+g'^2}\,ds$，理由是把曲线切成小段、用弦代替弧、再取极限；严格化需要导数的一致连续性。
• 单位圆给出 $\arccos x=\int_x^1\frac{ds}{\sqrt{1-s^2}}$，并由基本定理得到 $\arccos'(x)=-\frac{1}{\sqrt{1-x^2}}$。
• 逐点收敛：每个固定的 $x$ 收敛，$N$ 可以依赖 $x$。
• 一致收敛：同一个 $N$ 对所有 $x$ 有效；一致收敛蕴含逐点收敛。
• $x^n$ 在 $[0,1]$ 上逐点收敛到不连续函数，但不一致收敛；半高点 $x_n=(1/2)^{1/n}$ 暴露了误差始终为 $1/2$。
• Weierstrass M 判别法：若 $|f_n|\le M_n$ 且 $\sum M_n<\infty$，则 $\sum f_n$ 一致收敛。
• $\sum x^n/n!$ 在每个 $[-L,L]$ 上一致收敛；幂级数在 $[-L,L]$（$L<R$）上一致收敛，端点需单独讨论。

理解检验

1. 说明弧长公式为什么合理，并指出把「弦长之和」变成积分需要什么条件。

参考答案：对分割 $a=s_0<\cdots<s_n=b$，第 $i$ 段用弦长近似：

$$|\gamma(s_i)-\gamma(s_{i-1})|=\sqrt{f'(\xi_i)^2+g'(\eta_i)^2}\,\Delta s_i,$$

其中 $\xi_i,\eta_i$ 由中值定理给出，落在同一小段内。求和得到 Riemann 和。要让它收敛到 $\int\sqrt{f'^2+g'^2}\,ds$，需要 $\xi_i,\eta_i$ 与段内任意点的函数值差可以一致地变小，即 $\sqrt{f'^2+g'^2}$ 在 $[a,b]$ 上一致连续；由于 $f',g'$ 在紧区间上连续，这一条成立。若只有逐点连续性而没有紧性，弦长和的收敛就没有保证。

2. 计算单位圆上半部分 $\gamma(s)=(s,\sqrt{1-s^2})$ 的弧长微元，并推出 $\arccos$ 的积分表示与导数。

参考答案：$f'=1$，$g'=-s/\sqrt{1-s^2}$，故

$$f'^2+g'^2=1+\frac{s^2}{1-s^2}=\frac{1}{1-s^2},$$

弧长微元为 $ds/\sqrt{1-s^2}$。从横坐标 $x$ 到 $1$ 的弧长是 $\theta=\int_x^1 ds/\sqrt{1-s^2}$；几何上 $x=\cos\theta$，所以 $\theta=\arccos x$。求导时把积分改写为 $\int_0^1-\int_0^x$：第一项为常数，第二项由基本定理求导得 $1/\sqrt{1-x^2}$，故 $\arccos'(x)=-1/\sqrt{1-x^2}$。负号来自变量在积分下限。

3. 证明 $f_n(x)=x^n$ 在 $[0,1]$ 上逐点收敛但不一致收敛。

参考答案：逐点收敛：固定 $x<1$，$x^n\to0$；固定 $x=1$，$x^n=1$。所以极限是 $f(x)=0$（$x<1$）与 $f(1)=1$。不一致：每个 $f_n$ 连续，$f_n(0)=0<1/2<1=f_n(1)$，由介值定理存在 $x_n\in(0,1)$ 使 $x_n^n=1/2$。此时 $f(x_n)=0$，故 $|f_n(x_n)-f(x_n)|=1/2$。取 $\varepsilon=1/4$，任何 $N$ 之后都存在这样的 $x_n$，所以不存在对所有 $x$ 一致的 $N$。直观上，半高点 $x_n=(1/2)^{1/n}\to1$，误差始终是 $1/2$。

4. 证明一致收敛蕴含逐点收敛，并说明两个定义中 $N$ 的差别。

参考答案：设 $f_n\to f$ 一致收敛，给定 $x\in I$ 与 $\varepsilon>0$，取一致收敛定义中的 $N$；它只依赖 $\varepsilon$，对所有 $x$ 有效，因此对这个 $x$ 也有效，于是 $n>N$ 时 $|f_n(x)-f(x)|<\varepsilon$，即逐点收敛。差别在于：一致收敛的 $N=N(\varepsilon)$；逐点收敛的 $N=N(\varepsilon,x)$，允许随 $x$ 增大而增大。$x^n$ 的例子中，要让 $x^n<\varepsilon$ 需要 $n>\log\varepsilon/\log x$，当 $x\to1$ 时分母趋于 $0$，所需 $n$ 无界，所以找不到统一的 $N$。

5. 陈述并证明 Weierstrass M 判别法。

参考答案：设 $|f_n(x)|\le M_n$ 对一切 $x\in I$ 成立，且 $\sum M_n<\infty$。令 $S_n=\sum_{k=0}^n f_k$。对 $n>m$，

$$|S_n(x)-S_m(x)|\le\sum_{k=m+1}^{n}|f_k(x)|\le\sum_{k=m+1}^{n}M_k\le\sum_{k=m+1}^{\infty}M_k .$$

由于 $\sum M_k$ 收敛，给定 $\varepsilon$ 存在 $N$，使 $m\ge N$ 时尾部 $<\varepsilon$。对每个固定的 $x$，$(S_n(x))$ 是 Cauchy 数列，故收敛，极限记为 $S(x)$。在上式中令 $n\to\infty$ 得 $|S(x)-S_m(x)|\le\sum_{k=m+1}^\infty M_k<\varepsilon$，对一切 $x$ 成立且右端与 $x$ 无关，故 $S_m\to S$ 一致收敛。

6. 用 M 判别法证明 $\sum x^n/n!$ 在 $[-L,L]$ 上一致收敛；并说明幂级数在 $[-L,L]$（$L<R$）上为何一致收敛。

参考答案：在 $[-L,L]$ 上 $|x^n/n!|\le L^n/n!$，而 $\sum L^n/n!$ 收敛（比值判别法：$L/(n+1)\to0$），故 M 判别法给出一致收敛。对幂级数 $\sum a_nx^n$，若 $L<R=1/M$（$M=\limsup|a_n|^{1/n}$），则 $|a_nx^n|\le|a_n|L^n$，且

$$\limsup(|a_n|L^n)^{1/n}=ML=L/R<1,$$

由根值判别法 $\sum|a_n|L^n$ 收敛，再用 M 判别法。注意必须 $L<R$ 严格成立：端点处级数可能收敛，但一致性一般不成立，例如 $\sum x^n$ 在 $[0,1)$ 上不一致收敛。

与前后课的关系

弧长把积分用在几何上，而 $\arccos$ 的积分表示展示了基本定理的另一种用法：先由几何定义函数，再用求导得到它的导数。这与「先给公式再验证」的顺序相反，是分析里更常见的构造方式。

后半讲为下一讲做好了准备。有了 Weierstrass M 判别法，就可以证明幂级数在收敛区间内闭一致收敛；而一致收敛正是逐项积分与逐项求导的前提。下一讲会证明：一致收敛的函数列，其积分收敛到极限函数的积分；在导数也一致收敛的额外条件下，导数同样可以交换次序。这两条结论把本节的技术条件变成了真正的运算规则。

Two strands

The previous lecture ended with the arc length formula as a definition. This lecture justifies it: why is arc length $\int\sqrt{f'^2+g'^2}\,ds$, and what does the unit circle give? The answer includes the classical integral representation of $\arccos$.

The lecture then turns to a different subject: convergence of sequences of functions. A sequence of numbers has one notion of convergence; a sequence of functions has two:

• Pointwise convergence: for each fixed $x$, the sequence $f_n(x)$ converges.
• Uniform convergence: one $N$ works for all $x$ at once.

The difference looks technical but decides whether limits may be exchanged with integrals and derivatives. The main tool for detecting uniform convergence is the Weierstrass M-test.

Timeline

Arc length: the definition and its justification

Let

$$\gamma(s)=\big(f(s),g(s)\big), \qquad s\in[a,b],$$

be a plane curve with $f,g$ differentiable and derivatives continuous. Its arc length is defined as

$$L=\int_a^b\sqrt{f'(s)^2+g'(s)^2}\;ds .$$

Why this is reasonable. Take a partition $a=s_0<s_1<\cdots<s_n=b$. The curve splits into small pieces; the $i$-th piece runs from $\gamma(s_{i-1})$ to $\gamma(s_i)$. Approximate its length by the chord:

$$\big|\gamma(s_i)-\gamma(s_{i-1})\big| =\sqrt{\big(f(s_i)-f(s_{i-1})\big)^2+\big(g(s_i)-g(s_{i-1})\big)^2}.$$

Apply the mean value theorem to $f$ and to $g$ separately: there are $\xi_i,\eta_i\in(s_{i-1},s_i)$ with

$$f(s_i)-f(s_{i-1})=f'(\xi_i)\Delta s_i, \qquad g(s_i)-g(s_{i-1})=g'(\eta_i)\Delta s_i,$$

where $\Delta s_i=s_i-s_{i-1}$. Substituting,

$$\big|\gamma(s_i)-\gamma(s_{i-1})\big| =\sqrt{f'(\xi_i)^2+g'(\eta_i)^2}\;\Delta s_i .$$

Summing over $i$ produces a Riemann sum. Since $f',g'$ are continuous on a compact interval, they are uniformly continuous; for a fine enough partition, $\xi_i$ and $\eta_i$ lie in the same small piece, and the factor $\sqrt{f'(\xi_i)^2+g'(\eta_i)^2}$ differs from $\sqrt{f'(s)^2+g'(s)^2}$ by a uniformly small amount on that piece. Hence the Riemann sums converge to

$$\int_a^b\sqrt{f'(s)^2+g'(s)^2}\;ds .$$

So the formula comes from cutting the curve into pieces, replacing arcs by chords, and passing to the limit. The rigorous step uses uniform continuity — exactly the mechanism behind integrability of continuous functions.

Example 1: a parabolic arc

Take

$$\gamma(s)=(s,s^2), \qquad s\in[0,1].$$

Then $f'=1$ and $g'=2s$, so

$$L=\int_0^1\sqrt{1+4s^2}\;ds .$$

The integral can be evaluated by substituting $2s=\sinh t$ or $2s=\tan\theta$; the lecture only asks for the integral. The point is that the arc length formula turns a geometric question into a definite integral.

Example 2: the unit circle and arccos

The upper half of the unit circle $x^2+y^2=1$ can be parametrized by

$$\gamma(s)=\big(s,\sqrt{1-s^2}\big), \qquad s\in[-1,1].$$

Here $f(s)=s$ and $g(s)=\sqrt{1-s^2}$. Differentiating,

$$f'(s)=1, \qquad g'(s)=\frac{1}{2\sqrt{1-s^2}}\cdot(-2s)=-\frac{s}{\sqrt{1-s^2}} .$$

Hence

$$f'(s)^2+g'(s)^2 =1+\frac{s^2}{1-s^2} =\frac{1-s^2+s^2}{1-s^2} =\frac{1}{1-s^2} .$$

So the arc length element of the circle is $\dfrac{ds}{\sqrt{1-s^2}}$.

Take the point on the circle with horizontal coordinate $x$ and walk along the circle to $(1,0)$; call the arc length $\theta$. Since $s$ is the horizontal coordinate, $s$ runs from $x$ to $1$:

$$\theta=\int_x^1\frac{ds}{\sqrt{1-s^2}} .$$

Geometrically $\theta$ is the central angle subtended by this arc, and the horizontal coordinate is $x=\cos\theta$. Hence $\theta=\arccos x$, giving

$$\boxed{\ \arccos x=\int_x^1\frac{ds}{\sqrt{1-s^2}}\ }$$

The formula is not memorized; it follows from "angle equals arc length" combined with the arc length formula.

The derivative of arccos comes for free

The variable is in the lower limit. Rewrite using additivity:

$$\int_x^1\frac{ds}{\sqrt{1-s^2}} = \int_0^1\frac{ds}{\sqrt{1-s^2}} - \int_0^x\frac{ds}{\sqrt{1-s^2}} .$$

The first term is constant, with derivative zero; the second has derivative equal to the integrand at $x$ by the fundamental theorem. Therefore

$$\frac{d}{dx}\arccos x =0-\frac{1}{\sqrt{1-x^2}} =-\frac{1}{\sqrt{1-x^2}} .$$

The lecture emphasizes the sign: when the variable sits in the lower limit, a minus sign appears, because $\int_x^1=-\int_1^x$.

Pointwise convergence

Let $I$ be an interval, $(f_n)$ a sequence of functions on $I$, and $f$ a function on $I$. Say $f_n\to f$ pointwise if for every fixed $x\in I$,

$$\lim_{n\to\infty}f_n(x)=f(x).$$

In $\varepsilon$ language: for every $x\in I$ and every $\varepsilon>0$ there is $N$ such that $n>N$ implies $|f_n(x)-f(x)|<\varepsilon$. The $N$ may depend on $x$.

Example 1. On $[0,1]$ let $f_n(x)=x^n$ and

$$f(x)= \begin{cases} 0, & 0\le x<1,\\ 1, & x=1 . \end{cases}$$

For fixed $x<1$, $x^n\to0$; for $x=1$, $1^n=1$. So $f_n\to f$ pointwise. Every $f_n$ is continuous, but the limit $f$ is not — pointwise convergence does not preserve continuity.

Example 2. On $\mathbb R$ let

$$f_n(x)=\sum_{k=0}^{n}\frac{x^k}{k!} .$$

For each fixed $x$ these are partial sums of the exponential series, converging to $e^x$. So $f_n\to e^x$ pointwise.

Uniform convergence

Say $f_n\to f$ uniformly if for every $\varepsilon>0$ there is $N$ such that

$$n>N \implies |f_n(x)-f(x)|<\varepsilon \quad\text{for all }x\in I .$$

The only change: the same $N$ must work for every $x$.

Lemma. Uniform convergence implies pointwise convergence.

Proof. Given $x$ and $\varepsilon$, take the $N$ from the uniform definition. It works for all $x$, in particular for this one, so $n>N$ gives $|f_n(x)-f(x)|<\varepsilon$. That is pointwise convergence (with the extra information that $N$ does not depend on $x$). $\square$

The converse fails; the next example shows it.

Counterexample: $x^n$ on $[0,1]$

Figure：Pointwise vs. uniform quantifiers, and why xⁿ cannot enter the ε-band。

As above, $f_n(x)=x^n\to f$ pointwise, where $f(1)=1$ and $f(x)=0$ for $x<1$. But the convergence is not uniform.

Proof. Each $f_n$ is continuous, with $f_n(0)=0$ and $f_n(1)=1$. By the intermediate value theorem there is $x_n\in(0,1)$ with

$$f_n(x_n)=x_n^n=\frac12$$

(explicitly $x_n=(1/2)^{1/n}$, and $x_n\to1$). Since $x_n<1$, $f(x_n)=0$, so

$$|f_n(x_n)-f(x_n)|=\frac12 .$$

Take $\varepsilon=\frac14$. If convergence were uniform, some $N$ would make $|f_n(x)-f(x)|<\frac14$ for all $x$ whenever $n>N$. But at $x=x_n$ the gap is $\frac12>\frac14$, a contradiction. $\square$

Geometrically, every $f_n$ climbs from $0$ to $1$ and must pass through $\frac12$; that "half-height point" moves right toward $1$ as $n$ grows. The limit function is $0$ there, so the error stays at $\frac12$ no matter how large $n$ is. Pointwise convergence only promises that each fixed point eventually becomes accurate; it does not promise that the whole interval becomes accurate at once.

This also explains why the limit of $x^n$ is discontinuous: uniformity is exactly what would let continuity pass to the limit.

The Weierstrass M-test

Theorem. Let $(f_n)$ be a sequence of functions on an interval $I$, and suppose there are nonnegative numbers $M_n$ with

$$|f_n(x)|\le M_n \quad\text{for all }x\in I,$$

and $\sum_{n=0}^\infty M_n<\infty$. Then $\sum_{n=0}^\infty f_n(x)$ converges uniformly on $I$.

Proof. Let

$$S_n(x)=\sum_{k=0}^{n}f_k(x).$$

For $n>m$,

$$S_n(x)-S_m(x)=\sum_{k=m+1}^{n}f_k(x),$$

so by the triangle inequality and the hypothesis,

$$|S_n(x)-S_m(x)| \le\sum_{k=m+1}^{n}|f_k(x)| \le\sum_{k=m+1}^{n}M_k \le\sum_{k=m+1}^{\infty}M_k .$$

Since $\sum M_k$ converges, for every $\varepsilon>0$ there is $N$ such that $m\ge N$ implies the tail is below $\varepsilon$.

Step 1: pointwise convergence. For each fixed $x$, the inequality says $(S_n(x))$ is Cauchy, hence convergent; call the limit $S(x)$. This defines a function $S$.

Step 2: upgrade to uniformity. In

$$|S_n(x)-S_m(x)|\le\sum_{k=m+1}^{\infty}M_k,$$

let $n\to\infty$ (the left side tends to $|S(x)-S_m(x)|$):

$$|S(x)-S_m(x)|\le\sum_{k=m+1}^{\infty}M_k<\varepsilon \qquad(m\ge N).$$

The right side is independent of $x$ and the inequality holds for all $x$. That is precisely uniform convergence of $S_m\to S$. $\square$

The last step is the heart: the tail bound carries no $x$. That is the purpose of the "$M$" — a sequence of numbers, independent of $x$, controls the tails of a whole family of functions.

Application: the exponential series

On $[-L,L]$ consider

$$f_n(x)=\frac{x^n}{n!} .$$

Take

$$M_n=\frac{L^n}{n!} .$$

For $|x|\le L$,

$$|f_n(x)|=\frac{|x|^n}{n!}\le\frac{L^n}{n!}=M_n .$$

The series $\sum L^n/n!$ converges (ratio test: consecutive ratios are $L/(n+1)\to0$). By the M-test,

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

converges uniformly on $[-L,L]$ to $e^x$. Since $L$ is arbitrary, it converges uniformly on every bounded interval — but not on all of $\mathbb R$, for reasons the next lecture makes precise.

Power series and uniform convergence on compact subintervals

Let

$$\sum_{n=0}^{\infty}a_nx^n$$

be a power series and set

$$M=\limsup_{n\to\infty}|a_n|^{1/n}, \qquad R=\frac1M,$$

with the conventions $M=0\Rightarrow R=\infty$ and $M=\infty\Rightarrow R=0$. The number $R$ is the radius of convergence.

Known facts:

• the series converges absolutely for $|x|<R$;
• it diverges for $|x|>R$;
• at $|x|=R$ each endpoint must be checked separately; there is no general rule.

Theorem. If $0<L<R$, then $\sum a_nx^n$ converges uniformly on $[-L,L]$.

Proof. For $|x|\le L$,

$$|a_nx^n|\le|a_n|L^n .$$

Since $M=\limsup|a_n|^{1/n}=1/R$,

$$\limsup_{n\to\infty}\big(|a_n|L^n\big)^{1/n} =ML=\frac{L}{R}<1 .$$

The root test gives convergence of $\sum|a_n|L^n$. With $M_n=|a_n|L^n$, the M-test yields uniform convergence on $[-L,L]$. $\square$

The hypothesis $L<R$ is strict. At the endpoints $\pm R$ convergence may or may not hold, and uniformity generally fails. For instance $\sum x^n$ has radius $1$; it converges uniformly on $[-L,L]$ for every $L<1$, but not on $[0,1)$ — the same phenomenon as the $x^n$ counterexample.

The lecture notes that uniformity matters because it permits term-by-term integration and differentiation: if $f_n\to f$ uniformly, then

$$\lim_{n\to\infty}\int_a^b f_n=\int_a^b f,$$

and under suitable extra hypotheses also $\lim f_n'=f'$. That is the subject of the next lecture.

What this lecture establishes

• Arc length is $\int_a^b\sqrt{f'^2+g'^2}\,ds$, justified by cutting the curve, using chords, and passing to the limit; the rigorous step needs uniform continuity of the derivatives.
• The unit circle gives $\arccos x=\int_x^1\frac{ds}{\sqrt{1-s^2}}$, and the fundamental theorem gives $\arccos'(x)=-\frac{1}{\sqrt{1-x^2}}$.
• Pointwise convergence: each fixed $x$ converges, with $N$ allowed to depend on $x$.
• Uniform convergence: one $N$ for all $x$; it implies pointwise convergence.
• $x^n$ on $[0,1]$ converges pointwise to a discontinuous function but not uniformly; the half-height points $x_n=(1/2)^{1/n}$ keep the error at $1/2$.
• Weierstrass M-test: if $|f_n|\le M_n$ and $\sum M_n<\infty$, then $\sum f_n$ converges uniformly.
• $\sum x^n/n!$ converges uniformly on every $[-L,L]$; a power series converges uniformly on $[-L,L]$ when $L<R$, with endpoints needing separate treatment.

Comprehension check

1. Explain why the arc length formula is reasonable, and state what is needed to turn "sum of chord lengths" into an integral.

Reference answer: For a partition $a=s_0<\cdots<s_n=b$, the $i$-th piece is approximated by its chord:

$$|\gamma(s_i)-\gamma(s_{i-1})|=\sqrt{f'(\xi_i)^2+g'(\eta_i)^2}\,\Delta s_i,$$

where $\xi_i,\eta_i$ come from the mean value theorem and lie in the same small piece. Summing gives a Riemann sum. For it to converge to $\int\sqrt{f'^2+g'^2}\,ds$, the factor $\sqrt{f'^2+g'^2}$ must vary by a uniformly small amount across each piece, i.e. it must be uniformly continuous on $[a,b]$; this holds because $f'$ and $g'$ are continuous on a compact interval. Without compactness, convergence of the chord sums is not guaranteed.

2. Compute the arc length element of the upper unit circle and derive the integral representation and derivative of arccos.

Reference answer: With $f'=1$ and $g'=-s/\sqrt{1-s^2}$,

$$f'^2+g'^2=1+\frac{s^2}{1-s^2}=\frac{1}{1-s^2},$$

so the element is $ds/\sqrt{1-s^2}$. The arc from horizontal coordinate $x$ to $1$ has length $\theta=\int_x^1 ds/\sqrt{1-s^2}$; geometrically $x=\cos\theta$, hence $\theta=\arccos x$. For the derivative, write the integral as $\int_0^1-\int_0^x$: the first term is constant, the second differentiates to $1/\sqrt{1-x^2}$ by the fundamental theorem, so $\arccos'(x)=-1/\sqrt{1-x^2}$. The minus sign comes from the variable in the lower limit.

3. Prove that $f_n(x)=x^n$ converges pointwise but not uniformly on $[0,1]$.

Reference answer: Pointwise, $x^n\to0$ for fixed $x<1$ and $x^n=1$ for $x=1$, so the limit is $f(x)=0$ for $x<1$ and $f(1)=1$. For non-uniformity, each $f_n$ is continuous with $f_n(0)=0<\frac12<1=f_n(1)$, so the intermediate value theorem gives $x_n\in(0,1)$ with $x_n^n=\frac12$. Then $f(x_n)=0$, so $|f_n(x_n)-f(x_n)|=\frac12$. Taking $\varepsilon=\frac14$, after any $N$ such an $x_n$ still violates the uniform bound, so no single $N$ works. Intuitively, the half-height point $x_n=(1/2)^{1/n}$ tends to $1$ and the error stays at $\frac12$.

4. Prove that uniform convergence implies pointwise convergence, and contrast the two $N$'s.

Reference answer: Given $x\in I$ and $\varepsilon>0$, take $N$ from the uniform definition; it depends only on $\varepsilon$ and works for every $x$, hence for this $x$, so $n>N$ gives $|f_n(x)-f(x)|<\varepsilon$. The difference is that uniform convergence has $N=N(\varepsilon)$, while pointwise convergence allows $N=N(\varepsilon,x)$. In the $x^n$ example, $x^n<\varepsilon$ requires $n>\log\varepsilon/\log x$, and the required $n$ blows up as $x\to1$, so no uniform $N$ exists.

5. State and prove the Weierstrass M-test.

Reference answer: Suppose $|f_n(x)|\le M_n$ for all $x\in I$ and $\sum M_n<\infty$. With $S_n=\sum_{k=0}^n f_k$, for $n>m$,

$$|S_n(x)-S_m(x)|\le\sum_{k=m+1}^{n}|f_k(x)|\le\sum_{k=m+1}^{n}M_k\le\sum_{k=m+1}^{\infty}M_k .$$

Since $\sum M_k$ converges, for every $\varepsilon>0$ there is $N$ with tail below $\varepsilon$ once $m\ge N$. For each fixed $x$, $(S_n(x))$ is Cauchy, hence converges; call the limit $S(x)$. Letting $n\to\infty$ in the inequality gives $|S(x)-S_m(x)|\le\sum_{k=m+1}^\infty M_k<\varepsilon$ for all $x$, with a right side independent of $x$; this is uniform convergence.

6. Use the M-test to show $\sum x^n/n!$ converges uniformly on $[-L,L]$, and explain uniform convergence of a power series on $[-L,L]$ with $L<R$.

Reference answer: On $[-L,L]$, $|x^n/n!|\le L^n/n!$, and $\sum L^n/n!$ converges (ratio test, ratios $L/(n+1)\to0$), so the M-test applies. For $\sum a_nx^n$ with $L<R=1/M$ and $M=\limsup|a_n|^{1/n}$, we have $|a_nx^n|\le|a_n|L^n$ and

$$\limsup(|a_n|L^n)^{1/n}=ML=L/R<1,$$

so the root test gives convergence of $\sum|a_n|L^n$, and the M-test gives uniformity. The inequality $L<R$ must be strict: at the endpoints the series may converge but uniformity generally fails, as with $\sum x^n$ on $[0,1)$.

Relation to neighbouring lectures

Arc length applies the integral to geometry, while the arccos representation shows another use of the fundamental theorem: define a function geometrically, then differentiate to identify its derivative. This is the reverse of "state a formula, then verify it", and it is the more common construction in analysis.

The second half prepares the next lecture. With the M-test, power series converge uniformly on compact subintervals of the interval of convergence; uniform convergence is precisely the hypothesis needed for term-by-term integration and differentiation. The next lecture proves that uniformly convergent sequences may be integrated term by term, and that derivatives may also be exchanged under an additional uniform convergence hypothesis on the derivatives. Those results turn the technical conditions here into working rules.