幂级数的导数

hbghlyj

complex.pdf说到幂级数的导数有两个证明：一个是在分析二中，另一个是在黎曼积分中
找到了第二个证明：
Ben Green的讲义（2020）已将PDF内容放入帖子Riemann integral 中
这正是 §5.3. Power series and radius of convergence Theorem 5.2.
也可以直接进入原PDF的第27页

hbghlyj

找到了第一个证明（分析 II）Theorem 8.16
Let the real or complex power series $f(x):=\sum_{k=0}^{\infty} c_k x^k$ have radius of convergence $R \in(0, \infty]$. Then $f$ is differentiable in $\{x:|x|<R\}$ and $f^{\prime}$ is given by term-by-term differentiation:
$$
f^{\prime}(x)=\sum_{k=1}^{\infty} k c_k x^{k-1}
$$Proof. Fix $x_0 \in \mathbb{C}$ with $\left|x_0\right|<R$ and fix $\rho \in \mathbb{R}$ with $\left|x_0\right|<\rho<R$. By Lemma 8.15, $g(x):=\sum_{k=1}^{\infty} k c_k x^{k-1}$ has radius of convergence $R$ and hence $g\left(x_0\right)$ is well defined. We also observe, applying Lemma 8.15 again, that $\sum_{k=2}^{\infty} k(k-1) c_k x^{k-2}$ has ROC $R$, and so converges absolutely at $\rho<R$. In particular
$$
M:=\sum_{k=2}^{\infty} k(k-1)\left|c_k\right| \rho^{k-2}<\infty
$$
Now to show $f^{\prime}\left(x_0\right)=g\left(x_0\right)$ it is enough to bound
$$\tag{10}
\left|\frac{f(x)-f\left(x_0\right)}{x-x_0}-g\left(x_0\right)\right|=\left|\sum_{k=1}^{\infty} c_k\left(\frac{x^k-x_0^k}{x-x_0}-k x_0^{k-1}\right)\right|
$$
when $x$ is sufficiently close to $x_0$, so wlog $|x|<\rho$. Summing a geometric series we have
$$
\frac{x^k-x_0^k}{x-x_0}=x_0^{k-1}+x_0^{k-2} x+\cdots+x^{k-1}
$$so for $k=1$ we have $\frac{x^k-x_0^k}{x-x_0}=k x_0^{k-1}$ and for $k \geq 2$ we get
$$
\begin{aligned}
\frac{x^k-x_0^k}{x-x_0}-k x_0^{k-1} & =\left(x_0^{k-1}+x_0^{k-2} x+\cdots+x^{k-1}\right)-\left(x_0^{k-1}+x_0^{k-1}+\cdots+x_0^{k-1}\right) \\
& =x_0^{k-1}(1-1)+x_0^{k-2}\left(x-x_0\right)+x_0^{k-3}\left(x^2-x_0^2\right)+\cdots+\left(x^{k-1}-x_0^{k-1}\right) \\
& =\left(x-x_0\right) \cdot\left(x_0^{k-2}+x_0^{k-1}\left(x+x_0\right)+\cdots+\left(x^{k-2}+\cdots+x_0^{k-2}\right)\right)
\end{aligned}
$$
Hence if $\left|x_0\right|,|x|<\rho$ we have
$$
\left|\frac{x^k-x_0^k}{x-x_0}-k x_0^{k-1}\right| \leq\left|x-x_0\right|\left(\rho^{k-2}+2 \rho^{k-2}+\cdots+(k-1) \rho^{k-2}\right)=\left|x-x_0\right| \cdot \frac{k(k-1)}{2} \rho^{k-2}
$$
But then, by (10),
$$
\left|\frac{f(x)-f\left(x_0\right)}{x-x_0}-g\left(x_0\right)\right| \leq\left|x-x_0\right| \sum_{k=2}^{\infty} \frac{1}{2} k(k-1)\left|c_k\right| \rho^{k-2}=\frac{M}{2}\left|x-x_0\right| \rightarrow 0 \text { as } x \rightarrow x_0
$$$\square$

hbghlyj

我有一个问题：是否有可能 $\sum a_nx^n$ 的收敛范围小于 $\sum na_nx^{n-1}$ 的收敛范围?

Theorem 5.2.
$\ldots$
$\ldots$
Finally, it follows from (5.1) and Lemma 5.1 that the radius of convergence of the power series for $f'$ is at most $R_1$. Since $R_1 < R$ was arbitary, the radius of convergence of this power series is at most $R$, as claimed.

我们已经证明它们具有相同的收敛半径，但是是否可能存在 $f'$ 收敛但 $f$ 不收敛的边界点

Czhang271828

hbghlyj 发表于 2023-5-17 06:02
我有一个问题：是否有可能 $\sum a_nx^n$ 的收敛范围小于 $\sum na_nx^{n-1}$ 的收敛范围?

我们已经证明它 ...

若级数 $\sum na_nx^n$ 收敛, 则根据 A-D 判别法可知
\[
\sum (na_nx^n)\cdot \dfrac{1}{n}=\sum a_nx^n
\]
收敛.