还差一件事：求导能否与极限交换

上一讲的结论是：若 $f_n$ 可积且 $f_n\to f$ 一致，则 $f$ 可积，且

$$\int_a^b f_n\to\int_a^b f .$$

积分在一列函数一致收敛时表现良好。求导则完全不同：一致收敛本身不足以推出 $f_n'\to f'$。原因很直接——导数只看到函数在每一点的局部变化，而一致收敛控制的是函数值之间的整体距离。一个函数列可以在每一点都被压得很近，但斜率剧烈振荡。

课程没有给出反例，而是直接给出一个更强的充分条件：不仅函数收敛，导数也要一致收敛。在这个条件下，求导确实可以与极限交换。

时间轴

定理：导数一致收敛时的求导交换

定理。 设 $f_n$ 在区间 $(A,B)$ 上可导，$f_n'$ 连续。假设：

1. 存在 $x_0\in(A,B)$，使数列 $f_n(x_0)$ 收敛到某个数 $c$；
2. $f_n'$ 在 $(A,B)$ 上一致收敛到某个函数 $g$。

则存在函数 $f$，满足：

• $f$ 可导，$f'$ 连续；
• $f_n\to f$ 在每个有界子区间 $[a,b]\subset(A,B)$ 上一致；
• $f_n'\to f'$ 在同一子区间上一致。

第一步：构造极限函数

由「连续函数的一致极限连续」，$g$ 连续。

定义

$$f(x)=c+\int_{x_0}^{x}g(s)\,ds .$$

这个定义有两处值得说明。第一，$g$ 连续，所以积分存在。第二，为什么从 $c$ 开始积分？因为要保证 $f(x_0)=c$，而 $c$ 正是 $f_n(x_0)$ 的极限；这样 $f$ 才可能是 $f_n$ 的极限。

由微积分基本定理，$f$ 可导，且

$$f'(x)=g(x).$$

$g$ 连续，所以 $f'$ 连续。这就完成了结论的前半部分。

第二步：一致收敛

固定有界子区间 $[a,b]\subset(A,B)$，设 $x\in[a,b]$。由微积分基本定理，对每个 $n$ 有

$$f_n(x)=f_n(x_0)+\int_{x_0}^{x}f_n'(s)\,ds .$$

两式相减：

$$\begin{aligned} f_n(x)-f(x) &=\bigl(f_n(x_0)-c\bigr) +\int_{x_0}^{x}\bigl(f_n'(s)-g(s)\bigr)\,ds . \end{aligned}$$

取绝对值并用三角不等式：

$$|f_n(x)-f(x)| \le |f_n(x_0)-c| +\left|\int_{x_0}^{x}\bigl(f_n'(s)-g(s)\bigr)\,ds\right| .$$

把积分放缩成「长度 × 最大误差」。若 $x\ge x_0$，

$$\left|\int_{x_0}^{x}\bigl(f_n'(s)-g(s)\bigr)\,ds\right| \le (x-x_0)\,d(f_n',g) \le (b-a)\,d(f_n',g).$$

若 $x<x_0$，则

$$\left|\int_{x_0}^{x}\cdots\right| =\left|\int_{x}^{x_0}\cdots\right| \le (x_0-x)\,d(f_n',g) \le (b-a)\,d(f_n',g).$$

两种情况合起来，对一切 $x\in[a,b]$：

$$|f_n(x)-f(x)|\le |f_n(x_0)-c|+(b-a)\,d(f_n',g).$$

右端与 $x$ 无关。当 $n\to\infty$ 时，第一项趋于 $0$（条件 1），第二项趋于 $0$（条件 2，导数一致收敛）。所以对任意 $\varepsilon>0$，存在 $N$，使 $n\ge N$ 时右端小于 $\varepsilon$，且对所有 $x$ 同时成立。这就是 $f_n\to f$ 在 $[a,b]$ 上一致。

第三步：导数收敛是免费的

结论的第二条不用证：条件 2 已经说 $f_n'\to g$ 一致，而 $g=f'$，所以 $f_n'\to f'$ 一致。$\square$

注意条件的结构：导数的收敛被假设，函数的收敛被推出。这与积分定理的方向相反——那里函数的一致收敛足以推出积分的收敛；这里函数的一致收敛是结论，不是假设。

例子：指数函数满足 $E'=E$

定义

$$E_n(x)=\sum_{k=0}^{n}\frac{x^k}{k!}, \qquad E(x)=\sum_{k=0}^{\infty}\frac{x^k}{k!}.$$

逐项求导。 因为 $E_n$ 是多项式，可以逐项求导：

$$E_n'(x)=\sum_{k=1}^{n}\frac{k\,x^{k-1}}{k!} =\sum_{k=1}^{n}\frac{x^{k-1}}{(k-1)!} =\sum_{j=0}^{n-1}\frac{x^{j}}{j!} =E_{n-1}(x).$$

最后一步做了指标替换 $j=k-1$。所以

$$E_n'=E_{n-1}.$$

检查定理的条件。

• $E_n'=E_{n-1}$ 在任意 $[-L,L]$ 上一致收敛到 $E$（Weierstrass $M$ 判别法）。
• 在 $x_0=0$ 处，$E_n(0)=1$ 对每个 $n\ge0$ 成立，所以 $E_n(0)\to1$。
• $E_n'$ 连续（多项式）。

三条全部满足。于是定理给出：存在函数 $f$，在 $[-L,L]$ 上 $E_n\to f$ 一致，且 $E_n'\to f'$ 一致。

识别极限。 由上一讲，$E_n\to E$ 一致。一致收敛的极限唯一（每个点上的极限唯一），所以 $f=E$。另一方面，$E_n'=E_{n-1}$，而 $E_{n-1}\to E$ 一致，所以 $f'=E$。两句话合起来：

$$E'=E .$$

结论不只是「$E$ 可导」，而是「$E$ 的导数恰好是它自己」——这正是把 $e^x$ 定义为幂级数后最想得到的性质。

一般幂级数：逐项求导不改变收敛半径

设

$$f(x)=\sum_{n=0}^{\infty}a_nx^n$$

的收敛半径为

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

并按惯例处理 $M=0$ 与 $M=\infty$。逐项求导得到新级数

$$\sum_{n=1}^{\infty}n\,a_nx^{n-1}.$$

它的系数是 $b_{n-1}=n a_n$。计算它对应的 $M'$：

$$M'=\limsup_{n\to\infty}\bigl|n\,a_n\bigr|^{1/n} =\limsup_{n\to\infty}n^{1/n}\,|a_n|^{1/n}.$$

关键极限是 $n^{1/n}\to1$。这一点可以写成

$$n^{1/n}=e^{\frac{\log n}{n}}, \qquad \frac{\log n}{n}\to0,$$

所以 $n^{1/n}\to e^0=1$。于是

$$M'=\limsup_{n\to\infty}|a_n|^{1/n}=M .$$

收敛半径不变。

课程在这里回应了一个提问：求导后 $x$ 的幂次整体降了一次，用 $n$ 次方根是否还合适？答案是两种算法都对。因为收敛半径本质上是与几何级数比较得到的，把幂次整体平移不影响它；若坚持用 $n-1$ 次方根，得到的 $\limsup$ 也相同。差别只在书写，不在结果。

迭代到 $m$ 阶导数

既然一次求导不改变收敛半径，就可以重复。求导 $m$ 次后：

$$f^{(m)}(x)=\sum_{n=m}^{\infty}\frac{n!}{(n-m)!}\,a_n\,x^{n-m}.$$

系数里的连乘积展开是

$$\frac{n!}{(n-m)!} =n(n-1)(n-2)\cdots(n-m+1),$$

正好是逐次求导时依次落下来的因子。每一次求导都保持同一个收敛半径，所以这个级数的收敛半径仍是 $R$。

定理：幂级数在收敛区间内光滑

定理。 设幂级数的收敛半径为 $R>0$，$0<L<R$。则：

1. 部分和 $f_n(x)=\sum_{k=0}^{n}a_kx^k$ 在 $[-L,L]$ 上一致收敛到 $f$；
2. 对每个 $m\ge1$，$f_n$ 的 $m$ 阶导数在 $[-L,L]$ 上一致收敛到

$$\sum_{k=m}^{\infty}\frac{k!}{(k-m)!}\,a_k\,x^{k-m}.$$

因此 $f$ 在 $(-R,R)$ 内无穷次可导，且

$$f^{(m)}(x)=\sum_{k=m}^{\infty}\frac{k!}{(k-m)!}\,a_k\,x^{k-m}.$$

证明思路。 取 $m=1$。由 Weierstrass $M$ 判别法，$\sum_k a_kx^k$ 与 $\sum_k k a_kx^{k-1}$ 在 $[-L,L]$ 上分别一致收敛；也就是说，部分和 $f_n$ 与其导数 $f_n'$ 都一致收敛。在 $x_0=0$ 处，$f_n(0)=a_0$ 收敛。应用求导交换定理，得到 $f_n\to f$ 一致、$f_n'\to f'$ 一致，而 $f_n'$ 的极限正是逐项求导级数。对 $m$ 归纳即可。$\square$

收敛必须是「在每个 $[-L,L]$ 上」，$L<R$。在端点 $x=\pm R$ 处级数可能发散，逐项求导也可能失去一致性，所以定理只在内闭区间上成立。

逐项积分

同样的思路配上上一讲的积分定理。设 $a,b$ 都在收敛区间内部，$f(x)=\sum_k a_kx^k$。部分和 $f_n$ 是多项式，可以逐项积分：

$$\int_a^b f_n(x)\,dx =\sum_{k=0}^{n}a_k\int_a^b x^k\,dx .$$

因为 $f_n\to f$ 在包含 $[a,b]$ 的某个 $[-L,L]$ 上一致，积分交换定理给出

$$\int_a^b f(x)\,dx =\sum_{k=0}^{\infty}a_k\int_a^b x^k\,dx .$$

若取 $a=0$、$b=x$，每一项算出来是

$$\int_0^x t^k\,dt=\frac{x^{k+1}}{k+1},$$

于是得到一个漂亮的公式：

$$\int_0^x f(t)\,dt =\sum_{k=0}^{\infty}\frac{a_k}{k+1}x^{k+1}.$$

这也是一个幂级数，收敛半径不变。所以「积分一个幂级数」和「求导一个幂级数」一样，都是逐项操作。

什么是常微分方程

微分方程是含未知函数及其导数的方程。三个小例子：

例 1。 $f'(x)=x$。解是

$$f(x)=\frac{x^2}{2}+C,$$

因为常数的导数是 $0$，而 $(x^2/2)'=x$。所有解都长这样。

例 2。 $f'(x)=\bigl(f(x)-1\bigr)^2$。取

$$f(x)=1-\frac1x,$$

则 $f(x)-1=-1/x$，平方得 $1/x^2$；而 $f'(x)=1/x^2$。所以它确实是解。这个例子说明解不一定是初等函数的简单组合，也说明解可能只在部分区间上有定义（$x=0$ 处无定义）。

补充：课程转录把这一例听成了「$f'$ 减 $f$ 平方」之类的形式，按验算应为 $f'=(f-1)^2$；板书在过程中修正过一次符号，这个版本与修正后的板书一致。

例 3。 $f\,f'\,f''=0$。在每个点上至少有一个因子为零；由连续性，其中一个因子必须在整段区间上恒为零。于是解只有两类：$f'\equiv0$（常数）或 $f''\equiv0$（一次函数，常数也算）。所以全部解就是一次函数 $f(x)=ax+b$。

「常」微分方程（ordinary differential equation，ODE）指未知函数只依赖一个变量。若依赖多个变量、出现偏导数，就是偏微分方程。

本课研究的形状

未知函数记为 $y=y(x)$，方程与初值条件为

$$y'(x)=f\bigl(y(x)\bigr)+g(x), \qquad y(0)=a .$$

其中 $f$ 可导且 $f'$ 连续，$g$ 连续，$a$ 是给定实数。$f$ 只通过未知函数 $y$ 进入方程，$g$ 是已知的强制项或零阶项。因为 $f$ 可以任意非线性，这是非线性方程，一般写不出显式解。

要问两个问题：

• 存在性：给定 $a$，是否至少有一个解？
• 唯一性：若有两个解，是否必须相同？

Picard–Lindelöf 定理

定理。 设 $f$ 可导、$f'$ 连续，$g$ 连续，$a\in\mathbb R$。则存在 $\delta>0$，使得初值问题

$$y'=f(y)+g(x), \qquad y(0)=a$$

在区间 $(-\delta,\delta)$ 上有解；并且解在 $(-\delta,\delta)$ 上唯一。

要点在「局部」：结论只保证在原点附近的一个小小区间上有解。区间能有多长取决于 $f$ 与 $g$，一般不能延伸到整个 $\mathbb R$。下一讲会用具体例子说明，解可能在有限点处爆破，根本不可能延拓。

证明工具：压缩映射

定理的证明不直接构造解，而是把解写成一个映射的不动点。需要的工具是压缩映射定理。

定义。 设 $(X,d)$ 是度量空间，$T:X\to X$。若存在常数 $c<1$，使得

$$d\bigl(T(x),T(y)\bigr)\le c\,d(x,y) \quad\text{对一切 }x,y\in X$$

成立，则称 $T$ 是压缩映射。

压缩映射定理。 若 $(X,d)$ 是 Cauchy 完备的度量空间，$T:X\to X$ 是压缩映射，则 $T$ 有唯一不动点：存在唯一的 $x_0\in X$ 使 $T(x_0)=x_0$。

这个定理前面已经证明过，当时用在 $\mathbb R$ 上，为 Newton 法寻找零点。这里要用它的度量空间版本：空间取某个区间上的连续函数，距离取 sup 距离。

为什么能连起来。 把初值问题改写成积分方程

$$y(x)=a+\int_0^x\bigl[f(y(s))+g(s)\bigr]\,ds .$$

右边定义了一个映射 $T$：输入一个连续函数 $y$，输出另一个连续函数。若 $y$ 是不动点，则 $y=Ty$，求导（微积分基本定理）立刻给出 $y'=f(y)+g$，且 $y(0)=a$。于是「解微分方程」变成「找压缩映射的不动点」。

难点在于：这个 $T$ 在全体连续函数上不是压缩映射。必须把空间限制到一个足够小的闭球上，才能让压缩常数小于 $1$。这正是下一讲的全部技术工作。

本节结论

• 求导与极限交换需要导数一致收敛；仅有函数一致收敛不够。
• 交换定理的证明用积分重构 $f$，把误差拆成「端点误差 + 长度 × 导数误差」。
• 幂级数逐项求导不改变收敛半径，可逐项求导任意多次；逐项积分同样成立。
• $E_n'=E_{n-1}$ 与 $E_n(0)=1$ 推出 $E'=E$。
• 常微分方程 $y'=f(y)+g(x)$、$y(0)=a$ 在局部有唯一解（Picard–Lindelöf）。
• 证明工具是压缩映射定理；关键是把空间限制成一个闭球，使映射成为压缩。

理解检验

1. 为什么「$f_n\to f$ 一致」推不出「$f_n'\to f'$」？定理改用哪个假设绕过了这个困难？

参考答案：一致收敛控制的是函数值之差 $\sup_x|f_n-f|$，而导数由差商决定，差商涉及两个邻近点的值。函数值可以被压得很近，但每一点的局部斜率仍可能剧烈振荡，因此没有任何机制迫使 $f_n'$ 收敛，更不用说收敛到 $f'$。定理改为假设导数列 $f_n'$ 一致收敛到 $g$，再把 $f$ 用 $g$ 的积分构造出来，于是 $f'=g$ 直接成立，函数的一致收敛反而成了结论。

2. 在定理证明中，为什么把 $f$ 定义为 $c+\int_{x_0}^{x}g$，而不是 $\int_{x_0}^{x}g$？

参考答案：因为要匹配条件 $f_n(x_0)\to c$。取 $f(x)=c+\int_{x_0}^{x}g$ 时，由积分在一点上的值为零，$f(x_0)=c$，正是 $f_n(x_0)$ 的极限；若省掉 $c$，则 $f(x_0)=0$，极限函数在 $x_0$ 处的值与数列的极限不符，$f_n$ 就不可能一致收敛到它。

3. 证明 $E'=E$ 时，定理的三条假设分别由什么事实提供？

参考答案：（1）$E_n(x_0)$ 收敛：取 $x_0=0$，则 $E_n(0)=1$ 对所有 $n$ 成立，收敛到 $1$。（2）导数一致收敛：$E_n'=E_{n-1}$，而由 Weierstrass $M$ 判别法，$E_{n-1}\to E$ 在任意 $[-L,L]$ 上一致。（3）导数连续：$E_n'=E_{n-1}$ 是多项式，连续。定理于是给出极限函数可导，且导数等于 $E_{n-1}$ 的极限 $E$，即 $E'=E$。

4. 为什么幂级数逐项求导后收敛半径不变？请写出关键极限。

参考答案：原级数系数 $a_n$ 给出 $M=\limsup|a_n|^{1/n}$，新级数系数是 $n a_n$，对应

$$M'=\limsup\bigl|n a_n\bigr|^{1/n} =\limsup n^{1/n}|a_n|^{1/n}.$$

关键极限是 $n^{1/n}\to1$，因为 $n^{1/n}=e^{(\log n)/n}$ 且 $(\log n)/n\to0$。于是 $M'=M$，收敛半径 $R=1/M$ 不变。

5. 为什么把 $T(y)(x)=a+\int_0^x[f(y(s))+g(s)]\,ds$ 的不动点叫做原 ODE 的解？反过来为什么每个解也是不动点？

参考答案：若 $y=Ty$，则 $y(x)=a+\int_0^x[f(y(s))+g(s)]ds$。由微积分基本定理，右端可导，导数为 $f(y(x))+g(x)$；又因为在 $x=0$ 处积分为零，$y(0)=a$。所以 $y$ 满足 ODE 与初值。反过来，若 $y$ 可导且 $y'=f(y)+g$、$y(0)=a$，对 $y'$ 从 $0$ 到 $x$ 积分并用微积分基本定理，得到 $y(x)-y(0)=\int_0^x[f(y(s))+g(s)]ds$，即 $y=Ty$。两个方向都只用到微积分基本定理。

与前后课的关系

上一讲证明了积分与一致极限可交换；本讲补上求导的那一半，并因此得到幂级数的全部微积分性质。后半段转向常微分方程，把问题重写成积分算子的不动点，为下一讲的技术工作搭好框架。

下一讲会做三件事：定义闭球上的映射 $T$，证明 $T$ 把闭球映入自身，再证明 $T$ 在该闭球上是压缩映射；然后由压缩映射定理得到局部解的存在唯一性。再下一讲（标为期末复习）补完唯一性的全局部分，并用 $y'=y^2$ 的爆破说明「局部」不是技术上的保守，而是真实现象。

One thing left: can derivatives pass through a limit?

Last lecture's conclusion: if $f_n$ is integrable and $f_n\to f$ uniformly, then $f$ is integrable and

$$\int_a^b f_n\to\int_a^b f .$$

Integration behaves well under uniform convergence. Differentiation is completely different: uniform convergence by itself does not imply $f_n'\to f'$. The reason is straightforward — a derivative sees only the local behaviour near each point, while uniform convergence controls the overall distance between function values. A sequence of functions can be squeezed together at every point while the slopes oscillate wildly.

The course does not give a counterexample but instead supplies a stronger sufficient condition: not only the functions converge, but the derivatives converge uniformly. Under that condition, differentiation does commute with the limit.

Timeline

Theorem: differentiation commutes when derivatives converge uniformly

Theorem. Let $f_n$ be differentiable on $(A,B)$ with $f_n'$ continuous. Assume:

1. for some $x_0\in(A,B)$, the sequence $f_n(x_0)$ converges to a number $c$;
2. $f_n'$ converges uniformly on $(A,B)$ to a function $g$.

Then there is a function $f$ such that:

• $f$ is differentiable and $f'$ is continuous;
• $f_n\to f$ uniformly on each bounded subinterval $[a,b]\subset(A,B)$;
• $f_n'\to f'$ uniformly on the same subinterval.

Step 1: build the limit function

By "a uniform limit of continuous functions is continuous", $g$ is continuous.

Define

$$f(x)=c+\int_{x_0}^{x}g(s)\,ds .$$

Two points deserve comment. First, $g$ is continuous, so the integral exists. Second, why integrate starting from $c$? Because we need $f(x_0)=c$, and $c$ is precisely the limit of $f_n(x_0)$; only then can $f$ be the limit of $f_n$.

By the fundamental theorem of calculus, $f$ is differentiable and

$$f'(x)=g(x).$$

Since $g$ is continuous, so is $f'$. This gives the first half of the conclusion.

Step 2: uniform convergence

Fix a bounded subinterval $[a,b]\subset(A,B)$ and let $x\in[a,b]$. By the fundamental theorem of calculus, for every $n$,

$$f_n(x)=f_n(x_0)+\int_{x_0}^{x}f_n'(s)\,ds .$$

Subtract:

$$\begin{aligned} f_n(x)-f(x) &=\bigl(f_n(x_0)-c\bigr) +\int_{x_0}^{x}\bigl(f_n'(s)-g(s)\bigr)\,ds . \end{aligned}$$

Take absolute values and apply the triangle inequality:

$$|f_n(x)-f(x)| \le |f_n(x_0)-c| +\left|\int_{x_0}^{x}\bigl(f_n'(s)-g(s)\bigr)\,ds\right| .$$

Bound the integral by "length times maximum error". If $x\ge x_0$,

$$\left|\int_{x_0}^{x}\bigl(f_n'(s)-g(s)\bigr)\,ds\right| \le (x-x_0)\,d(f_n',g) \le (b-a)\,d(f_n',g).$$

If $x<x_0$, then

$$\left|\int_{x_0}^{x}\cdots\right| =\left|\int_{x}^{x_0}\cdots\right| \le (x_0-x)\,d(f_n',g) \le (b-a)\,d(f_n',g).$$

In either case, for all $x\in[a,b]$,

$$|f_n(x)-f(x)|\le |f_n(x_0)-c|+(b-a)\,d(f_n',g).$$

The right side is independent of $x$. As $n\to\infty$ the first term tends to $0$ (hypothesis 1) and the second to $0$ (hypothesis 2, uniform convergence of derivatives). So for every $\varepsilon>0$ there is $N$ such that for $n\ge N$ the right side is below $\varepsilon$, simultaneously for all $x$. That is uniform convergence of $f_n$ to $f$ on $[a,b]$.

Step 3: derivative convergence is free

The second conclusion needs no work: hypothesis 2 already says $f_n'\to g$ uniformly, and $g=f'$. Hence $f_n'\to f'$ uniformly. $\square$

Note the structure of the hypotheses: convergence of the derivatives is assumed; convergence of the functions is derived. This is the opposite direction from the integration theorem, where uniform convergence of the functions implied convergence of the integrals.

Example: the exponential satisfies $E'=E$

Define

$$E_n(x)=\sum_{k=0}^{n}\frac{x^k}{k!}, \qquad E(x)=\sum_{k=0}^{\infty}\frac{x^k}{k!}.$$

Differentiate term by term. Since $E_n$ is a polynomial,

$$E_n'(x)=\sum_{k=1}^{n}\frac{k\,x^{k-1}}{k!} =\sum_{k=1}^{n}\frac{x^{k-1}}{(k-1)!} =\sum_{j=0}^{n-1}\frac{x^{j}}{j!} =E_{n-1}(x),$$

where the last step is the index shift $j=k-1$. Hence

$$E_n'=E_{n-1}.$$

Check the hypotheses.

• $E_n'=E_{n-1}$ converges uniformly to $E$ on every $[-L,L]$ (Weierstrass $M$-test).
• At $x_0=0$, $E_n(0)=1$ for every $n\ge0$, so $E_n(0)\to1$.
• $E_n'$ is continuous (a polynomial).

All three hold. The theorem gives a function $f$ with $E_n\to f$ uniformly and $E_n'\to f'$ uniformly on $[-L,L]$.

Identify the limit. By the previous lecture, $E_n\to E$ uniformly. A uniform limit is unique (each point has a unique limit), so $f=E$. On the other hand $E_n'=E_{n-1}$ and $E_{n-1}\to E$ uniformly, so $f'=E$. Combining:

$$E'=E .$$

The conclusion is not merely that $E$ is differentiable, but that its derivative is itself — exactly the property wanted when $e^x$ is defined by its power series.

General power series: differentiation preserves the radius

Let

$$f(x)=\sum_{n=0}^{\infty}a_nx^n$$

with radius of convergence

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

with the usual conventions for $M=0$ and $M=\infty$. Term-by-term differentiation gives

$$\sum_{n=1}^{\infty}n\,a_nx^{n-1}.$$

Its coefficients are $b_{n-1}=n a_n$. Its $M'$ is

$$M'=\limsup_{n\to\infty}\bigl|n\,a_n\bigr|^{1/n} =\limsup_{n\to\infty}n^{1/n}\,|a_n|^{1/n}.$$

The key limit is $n^{1/n}\to1$:

$$n^{1/n}=e^{\frac{\log n}{n}}, \qquad \frac{\log n}{n}\to0,$$

so $n^{1/n}\to e^0=1$. Hence

$$M'=\limsup_{n\to\infty}|a_n|^{1/n}=M .$$

The radius is unchanged.

A student asked whether the $n$-th root is still the right tool after the powers shift down by one. The answer: both computations work. The radius comes from comparison with a geometric series, and shifting all powers by one does not affect it; using the $(n-1)$-st root gives the same limsup. The difference is bookkeeping, not substance.

Iterating to the $m$-th derivative

Since one differentiation preserves the radius, repeat. After $m$ differentiations,

$$f^{(m)}(x)=\sum_{n=m}^{\infty}\frac{n!}{(n-m)!}\,a_n\,x^{n-m}.$$

The coefficient expands as

$$\frac{n!}{(n-m)!} =n(n-1)(n-2)\cdots(n-m+1),$$

the factors that drop down at each differentiation. Each step preserves the same radius, so this series again has radius $R$.

Theorem: power series are smooth inside the radius

Theorem. Let a power series have radius $R>0$ and let $0<L<R$. Then:

1. the partial sums $f_n(x)=\sum_{k=0}^{n}a_kx^k$ converge uniformly to $f$ on $[-L,L]$;
2. for every $m\ge1$, the $m$-th derivatives of $f_n$ converge uniformly on $[-L,L]$ to

$$\sum_{k=m}^{\infty}\frac{k!}{(k-m)!}\,a_k\,x^{k-m}.$$

Consequently $f$ is infinitely differentiable on $(-R,R)$ and

$$f^{(m)}(x)=\sum_{k=m}^{\infty}\frac{k!}{(k-m)!}\,a_k\,x^{k-m}.$$

Idea of the proof. Take $m=1$. By the Weierstrass $M$-test, both $\sum_k a_kx^k$ and $\sum_k k a_kx^{k-1}$ converge uniformly on $[-L,L]$; that is, the partial sums $f_n$ and their derivatives $f_n'$ both converge uniformly. At $x_0=0$, $f_n(0)=a_0$ converges. The differentiation theorem then gives $f_n\to f$ and $f_n'\to f'$ uniformly, and the limit of $f_n'$ is the term-by-term derivative. Induct on $m$. $\square$

The convergence is on each $[-L,L]$ with $L<R$. At the endpoints $x=\pm R$ the series may diverge, and the differentiated series may lose uniformity, so the theorem is stated on inner closed intervals only.

Term-by-term integration

The same idea with the previous lecture's integration theorem. Let $a,b$ lie inside the interval of convergence and $f(x)=\sum_k a_kx^k$. The partial sums are polynomials, so

$$\int_a^b f_n(x)\,dx =\sum_{k=0}^{n}a_k\int_a^b x^k\,dx .$$

Since $f_n\to f$ uniformly on some $[-L,L]$ containing $[a,b]$, the interchange theorem gives

$$\int_a^b f(x)\,dx =\sum_{k=0}^{\infty}a_k\int_a^b x^k\,dx .$$

Taking $a=0$, $b=x$, each term is

$$\int_0^x t^k\,dt=\frac{x^{k+1}}{k+1},$$

so

$$\int_0^x f(t)\,dt =\sum_{k=0}^{\infty}\frac{a_k}{k+1}x^{k+1}.$$

This is again a power series with the same radius. Integrating a power series, like differentiating one, is a term-by-term operation.

What an ordinary differential equation is

A differential equation is an equation involving an unknown function and its derivative. Three small examples.

Example 1. $f'(x)=x$. The solutions are

$$f(x)=\frac{x^2}{2}+C,$$

because the derivative of a constant is $0$ and $(x^2/2)'=x$. These are all the solutions.

Example 2. $f'(x)=\bigl(f(x)-1\bigr)^2$. Take

$$f(x)=1-\frac1x .$$

Then $f(x)-1=-1/x$, whose square is $1/x^2$, while $f'(x)=1/x^2$. So it solves the equation. This example shows that solutions need not be simple combinations of elementary functions, and that a solution may be defined only on part of the real line (it is undefined at $x=0$).

Note: the lecture transcript mishears this example as something like "$f'$ minus $f$ squared"; checking the algebra, the equation is $f'=(f-1)^2$, matching the board after the instructor corrected a sign in the middle.

Example 3. $f\,f'\,f''=0$. At every point at least one factor vanishes; by continuity one of them must vanish identically on the whole interval. So the solutions fall into two classes: $f'\equiv0$ (constants) or $f''\equiv0$ (linear functions, constants included). All solutions are therefore linear, $f(x)=ax+b$.

An ordinary differential equation (ODE) is one whose unknown depends on a single variable. If it depends on several variables and partial derivatives appear, it is a partial differential equation.

The shape studied in this course

Write the unknown as $y=y(x)$ and consider

$$y'(x)=f\bigl(y(x)\bigr)+g(x), \qquad y(0)=a ,$$

where $f$ is differentiable with $f'$ continuous, $g$ is continuous, and $a$ is a given real number. The unknown enters only through $y$ inside $f$, while $g$ is a known forcing term. Since $f$ may be any nonlinearity, this is a nonlinear equation, and explicit solutions generally do not exist.

Two questions:

• Existence: given $a$, is there at least one solution?
• Uniqueness: if there are two, must they coincide?

The Picard–Lindelöf theorem

Theorem. Let $f$ be differentiable with $f'$ continuous, let $g$ be continuous, and let $a\in\mathbb R$. Then there is $\delta>0$ such that the initial value problem

$$y'=f(y)+g(x), \qquad y(0)=a$$

has a solution on $(-\delta,\delta)$, and the solution on $(-\delta,\delta)$ is unique.

The essential word is local: the conclusion is only that a solution exists on a small interval around the origin. How large the interval can be depends on $f$ and $g$, and in general it cannot be extended to all of $\mathbb R$. The next lecture gives an example where the solution blows up at a finite point and cannot be extended.

The tool: contraction mappings

The proof does not construct the solution directly. Instead it writes the solution as a fixed point of a map.

Definition. Let $(X,d)$ be a metric space and $T:X\to X$. If there is a constant $c<1$ such that

$$d\bigl(T(x),T(y)\bigr)\le c\,d(x,y) \quad\text{for all }x,y\in X,$$

then $T$ is a contraction.

Contraction mapping theorem. If $(X,d)$ is Cauchy complete and $T:X\to X$ is a contraction, then $T$ has a unique fixed point: there is exactly one $x_0\in X$ with $T(x_0)=x_0$.

This theorem was proved earlier, on $\mathbb R$, for Newton's method of finding zeros. Here its metric-space version is needed: the space is a set of continuous functions on an interval, and the distance is the sup distance.

How they connect. Rewrite the initial value problem as the integral equation

$$y(x)=a+\int_0^x\bigl[f(y(s))+g(s)\bigr]\,ds .$$

The right side defines a map $T$: it takes a continuous function $y$ to another continuous function. If $y$ is a fixed point, then $y=Ty$, and differentiating (fundamental theorem of calculus) gives $y'=f(y)+g$, while $y(0)=a$. Thus "solve the ODE" becomes "find a fixed point of a contraction".

The difficulty: this $T$ is not a contraction on the whole space of continuous functions. The space must be restricted to a sufficiently small closed ball before the contraction constant can be made less than $1$. That restriction is the entire technical content of the next lecture.

What this lecture establishes

• Differentiation commutes with limits when the derivatives converge uniformly; uniform convergence of the functions alone is not enough.
• The proof reconstructs $f$ by integration and splits the error into an endpoint term plus "length times derivative error".
• Term-by-term differentiation of a power series preserves the radius and may be repeated indefinitely; term-by-term integration holds as well.
• $E_n'=E_{n-1}$ together with $E_n(0)=1$ yields $E'=E$.
• The ODE $y'=f(y)+g(x)$, $y(0)=a$ has a unique local solution (Picard–Lindelöf).
• The proof uses the contraction mapping theorem; the key step restricts the space to a closed ball so that the map becomes a contraction.

Comprehension check

1. Why does uniform convergence of $f_n$ fail to imply $f_n'\to f'$? Which hypothesis in the theorem avoids the difficulty?

Reference answer: Uniform convergence controls $\sup_x|f_n-f|$, i.e. differences of values, whereas derivatives are determined by difference quotients involving two nearby points. Values can be squeezed together while the local slope at each point still oscillates; nothing forces $f_n'$ to converge, let alone to $f'$. The theorem instead assumes that the derivatives $f_n'$ converge uniformly to $g$, and reconstructs $f$ as the integral of $g$, so $f'=g$ holds by construction and uniform convergence of the functions becomes the conclusion rather than an assumption.

2. In the proof, why define $f=c+\int_{x_0}^{x}g$ rather than just $\int_{x_0}^{x}g$?

Reference answer: To match the hypothesis $f_n(x_0)\to c$. With $f(x)=c+\int_{x_0}^{x}g$, the integral over a single point is zero, so $f(x_0)=c$, exactly the limit of $f_n(x_0)$. Without the constant, $f(x_0)=0$, which generally differs from the limit of the values at $x_0$, and $f_n$ could not converge uniformly to it.

3. In proving $E'=E$, which facts supply the three hypotheses of the theorem?

Reference answer: (1) Convergence at a point: take $x_0=0$, then $E_n(0)=1$ for all $n$, so it converges to $1$. (2) Uniform convergence of derivatives: $E_n'=E_{n-1}$, and by the Weierstrass $M$-test $E_{n-1}\to E$ uniformly on every $[-L,L]$. (3) Continuity of derivatives: $E_n'=E_{n-1}$ is a polynomial. The theorem then gives a differentiable limit with derivative equal to the limit of $E_{n-1}$, namely $E$; hence $E'=E$.

4. Why is the radius of convergence unchanged by term-by-term differentiation? Write the key limit.

Reference answer: The original coefficients give $M=\limsup|a_n|^{1/n}$; the differentiated series has coefficients $n a_n$, so

$$M'=\limsup\bigl|n a_n\bigr|^{1/n} =\limsup n^{1/n}|a_n|^{1/n}.$$

The key limit is $n^{1/n}\to1$, since $n^{1/n}=e^{(\log n)/n}$ and $(\log n)/n\to0$. Hence $M'=M$ and the radius $R=1/M$ is unchanged.

5. Why is a fixed point of $T(y)(x)=a+\int_0^x[f(y(s))+g(s)]ds$ a solution of the ODE, and why is every solution a fixed point?

Reference answer: If $y=Ty$, then $y(x)=a+\int_0^x[f(y(s))+g(s)]ds$. By the fundamental theorem of calculus the right side is differentiable with derivative $f(y(x))+g(x)$; and since the integral over $[0,0]$ vanishes, $y(0)=a$. So $y$ satisfies the ODE and the initial condition. Conversely, if $y$ is differentiable with $y'=f(y)+g$ and $y(0)=a$, integrating $y'$ from $0$ to $x$ and using the fundamental theorem gives $y(x)-y(0)=\int_0^x[f(y(s))+g(s)]ds$, i.e. $y=Ty$. Both directions use only the fundamental theorem of calculus.

Relation to neighbouring lectures

The previous lecture proved that integration commutes with uniform limits; this lecture supplies the differentiation half and thereby the full calculus of power series. The second half turns to ODEs, rewriting the problem as a fixed point of an integral operator and setting up the next lecture's technical work.

That lecture does three things: define the map $T$ on a closed ball, prove that $T$ maps the ball into itself, and prove that $T$ is a contraction there; the contraction mapping theorem then yields local existence and uniqueness. The following lecture (labelled a final review) completes the global uniqueness argument and uses the blow-up of $y'=y^2$ to show that "local" is not technical caution but a real phenomenon.