为什么对e^x微分的时候不需要用 chain rule

foxupskirt

想了很久也没有想明白

foxupskirt

hbghlyj 发表于 2023-5-8 15:55
d(a^x)/dx = a^x ln a
当 a = e 时得到
d(e^x)/dx = e^x

对，你说得没错，但这个是结果，我想知道的是那个过程。🙏🏻🙏🏻

hbghlyj

foxupskirt 发表于 2023-5-8 09:01
对，你说得没错，但这个是结果，我想知道的是那个过程。🙏🏻🙏🏻

根据定义$\displaystyle \exp z:=\sum _{k=0}^{\infty }{\frac {z^{k}}{k!}}$
我们假设$\rmd\over\rmd z$与$∑$可以交换
$$\frac{\rmd}{\rmd z}\exp z=\sum _{k=1}^{\infty }\frac{\rmd}{\rmd z}{\frac {z^{k}}{k!}}=\sum _{k=1}^{\infty }{\frac {z^{k-1}}{(k-1)!}}=\exp z$$

hbghlyj

foxupskirt 发表于 2023-5-8 09:01
对，你说得没错，但这个是结果，我想知道的是那个过程。🙏🏻🙏🏻

根据定义$\exp z=\lim_{n\to\infty}\left(1+\frac zn\right)^n$
$$\frac{\rmd}{\rmd z}\exp z=\frac{\rmd}{\rmd z}\lim_{n\to\infty}\left(1+\frac zn\right)^n$$
我们假设$\rmd\over\rmd z$与$\lim$可以交换
$$\frac{\rmd}{\rmd z}\exp z=\lim_{n\to\infty}\frac{\rmd}{\rmd z}\left(1+\frac zn\right)^n\xlongequal{\text{chain rule}}\lim_{n\to\infty}\frac1n\cdot n\left(1+\frac zn\right)^{n-1}=\exp z$$

hbghlyj

hbghlyj 发表于 2023-5-8 10:15
我们假设$\rmd\over\rmd z$与$\lim$可以交换

Analysis I by Herbert Amann, Page 373 函数列的极限的可导性
2.8 Theorem (differentiability of the limits of sequences of functions) Let $X$ be an open (or convex) perfect subset of $\mathbb{K}$ and $f_n \in C^1(X, E)$ for all $n \in \mathbb{N}$. Suppose that there are $f, g \in E^X$ such that
(i) $\left(f_n\right)$ converges pointwise to $f$, and
(ii) $\left(f_n^{\prime}\right)$ converges locally uniformly to $g$.
Then $f$ is in $C^1(X, E)$, and $f^{\prime}=g$. In addition, $\left(f_n\right)$ converges locally uniformly to $f$.

Proof. Let $a \in X$. Then there is some $r>0$ such that $\left(f_n^{\prime}\right)$ converges uniformly to $g$ on $B_r:=\mathbb{B}_{\mathbb{K}}(a, r) \cap X$. If $X$ is open we can choose $r>0$ so that $\mathbb{B}(a, r)$ is contained in $X$. Hence with either of our assumptions, $B_r$ is convex and perfect. Thus, for each $x \in B_r$, we can apply the mean value theorem (Theorem IV.2.18) to the function
$$
[0,1] \rightarrow E, \quad t \mapsto f_n(a+t(x-a))-t f_n^{\prime}(a)(x-a)
$$
to get
$$
\left|f_n(x)-f_n(a)-f_n^{\prime}(a)(x-a)\right| \leq \sup _{0<t<1}\left|f_n^{\prime}(a+t(x-a))-f_n^{\prime}(a)\right||x-a|
$$
Taking the limit $n \rightarrow \infty$ we get
$$\tag{2.1}\label1
|f(x)-f(a)-g(a)(x-a)| \leq \sup _{0<t<1}|g(a+t(x-a))-g(a)||x-a|
$$
for each $x \in B_r$. Theorem 2.4 shows that $g$ is in $C(X, E)$, so it follows from \eqref{1} that
$$
f(x)-f(a)-g(a)(x-a)=o(|x-a|) \quad(x \rightarrow a) .
$$
Hence $f$ is differentiable at $a$ and $f^{\prime}(a)=g(a)$. Since this holds for all $a \in X$, we have shown that $f \in C^1(X, E)$.

It remains to prove that $\left(f_n\right)$ converges locally uniformly to $f$. Applying the mean value theorem to the function
$$
[0,1] \rightarrow E, \quad t \mapsto\left(f_n-f\right)(a+t(x-a))
$$
we get the inequality
$$
\begin{aligned}
\left|f_n(x)-f(x)\right| & \leq\left|f_n(x)-f(x)-\left(f_n(a)-f(a)\right)\right|+\left|f_n(a)-f(a)\right| \\
& \leq r \sup _{0<t<1}\left|f_n^{\prime}(a+t(x-a))-f^{\prime}(a+t(x-a))\right|+\left|f_n(a)-f(a)\right| \\
& \leq r\left\|f_n^{\prime}-f^{\prime}\right\|_{\infty, B_r}+\left|f_n(a)-f(a)\right|
\end{aligned}
$$
for each $x \in B_r$. The right side of this inequality is independent of $x \in B_r$ and converges to 0 as $n \rightarrow \infty$ because of (ii) and the fact that $f^{\prime}=g$. Thus $\left(f_n\right)$ converges uniformly to $f$ on $B_r$.  $_\blacksquare$

---

2.9 Corollary (differentiability of the limit of a series of functions) Suppose that $X \subseteq \mathbb{K}$ is open (or convex) and perfect, and $\left(f_n\right)$ is a sequence in $C^1(X, E)$ for which $\sum_{f_n} f_n$ converges pointwise and $\sum f_n^{\prime}$ converges locally uniformly. Then the $\operatorname{sum} \sum_{n=0}^{\infty} f_n$ is in $C^1(X, E)$ and
$$
\left(\sum_{n=0}^{\infty} f_n\right)^{\prime}=\sum_{n=0}^{\infty} f_n^{\prime} .
$$
In addition, $\sum f_n$ converges locally uniformly.
Proof This follows directly from Theorem 2.8.  $_\blacksquare$

---

2.10 Remarks (a) Let $\left(f_n\right)$ be a sequence in $C^1(X, E)$ which converges uniformly to $f$. Even if $f$ is continuously differentiable, $\left(f_n^{\prime}\right)$ does not, in general, converge pointwise to $f^{\prime}$.
Proof Let $X:=\mathbb{R}, E:=\mathbb{R}$ and $f_n(x):=(1 / n) \sin (n x)$ for all $n \in \mathbb{N}^{\times}$. Because
$$
\left|f_n(x)\right|=|\sin (n x)| / n \leq 1 / n, \quad x \in X,
$$
$\left(f_n\right)$ converges uniformly to 0 . Since $\lim f_n^{\prime}(0)=1$, the sequence $\left(f_n^{\prime}(0)\right)$ does not converge to the derivative of the limit function at the point 0.  $_\blacksquare$
(b) Let $\left(f_n\right)$ be a sequence in $C^1(X, E)$ such that $\sum f_n$ converges uniformly. Then, in general, $\sum f_n^{\prime}$ does not converge even pointwise.
Proof Suppose that $X:=\mathbb{R}, E:=\mathbb{R}$, and $f_n(x):=\left(1 / n^2\right) \sin (n x)$ for all $n \in \mathbb{N}^{\times}$. Then $\left\|f_n\right\|_{\infty}=1 / n^2$ and so, by the Weierstrass majorant criterion, the series $\sum f_n$ converges uniformly. Since $f_n^{\prime}(x)=(1 / n) \cos (n x), \sum f_n^{\prime}(0)$ does not converge.  $_\blacksquare$

hbghlyj

hbghlyj 发表于 2023-5-8 10:11
我们假设$\rmd\over\rmd z$与$∑$可以交换

Analysis I by Herbert Amann, Page 377 幂级数的可导性
Differentiability of Power Series
Let $a=\sum_k a_k X^k \in \mathbb{K} \llbracket X \rrbracket$ be a power series with radius of convergence $\rho=\rho_a>0$, and $\underline{a}$ the function on $\rho \mathbb{B}_{\mathbb{K}}$ represented by $a$. When no misunderstanding is possible, we write $\mathbb{B}$ for $\mathbb{B}_{\mathbb{K}}$.
3.1 Theorem (differentiability of power series) Let $a=\sum_k a_k X^k$ be a power series. Then $\underline{a}$ is continuously differentiable on $\rho \mathbb{B}$. The

series $\sum_{k \geq 1} k a_k X^{k-1}$ has radius of convergence $\rho$ and
$$
\underline{a}^{\prime}(x)=\left(\sum_{k=0}^{\infty} a_k x^k\right)^{\prime}=\sum_{k=1}^{\infty} k a_k x^{k-1}, \quad x \in \rho \mathbb{B}
$$
Proof Let $\rho^{\prime}$ be the radius of convergence of the power series $\sum k a_k X^{k-1}$. From Hadamard's formula (II.9.3), Example II.4.2(d) and Exercise II.5.2(d) we have
$$
\rho^{\prime}=\frac{1}{\overline{\lim } \sqrt[k]{k\left|a_k\right|}}=\frac{1}{\lim \sqrt[k]{k} \overline{\lim } \sqrt[k]{\left|a_k\right|}}=\frac{1}{\overline{\lim } \sqrt[k]{\left|a_k\right|}}=\rho .
$$
By Theorem 1.8, the power series $\sum_{k \geq 1} k a_k X^{k-1}$ converges locally uniformly on $\rho \mathbb{B}$, so the claim follows from Corollary 2.9.  $_\blacksquare$

---

3.2 Corollary If $a=\sum a_k X^k$ is a power series with positive radius of convergence $\rho$, then $\underline{a} \in C^{\infty}(\rho \mathbb{B}, \mathbb{K})$ and $\underline{a}=\mathcal{T}(\underline{a}, 0)$. In other words, $\sum a_k X^k$ is the Taylor series of $\underline{a}$ at 0 and $a_k=\underline{a}^{(k)}(0) / k !$.

Proof By induction, it follows from Theorem 3.1, that $\underline{a}$ is smooth on $\rho \mathbb{B}$ and that, for all $x \in \rho \mathbb{B}$,
$$
\underline{a}^{(k)}(x)=\sum_{n=k}^{\infty} n(n-1) \cdots(n-k+1) a_n x^{n-k}, \quad k \in \mathbb{N} .
$$
Hence $\underline{a}^{(k)}(0)=k ! a_k$ for all $k \in \mathbb{N}$ and we have proved the claim.  $_\blacksquare$

foxupskirt

hbghlyj 发表于 2023-5-8 17:23
Analysis I by Herbert Amann, Page 373 函数列的极限的可导性
2.8 Theorem (differentiability ...

好的，谢谢您，我先研究一下

Czhang271828

chain rule 对复合函数用, 明显 $e^x$ 是个简单函数, 不必复杂化. 这个问题和 "为什么给 $f(x)=Cx$ 求导不用 chain rule" 差不多, 不知道题主希望看到什么样的答案?

hbghlyj

${\rmd\over\rmd x}(\exp(\exp(\exp(x)))) = \exp(x + \exp(x)+ \exp(\exp(x)) )$