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qedhere是个好东西

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2022-11-07 11:08:25 +08:00
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03函数的导数.tex
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@@ -54,7 +54,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
\begin{proof}[解]
对于任意的$x$，有
\[\deriv{f}(x) = \tolim{\delx}{0}\frac{f(x + \delx) - f(x)}{\delx} = \tolim{\delx}{0} \frac{c - c}{\delx} = 0 \eqper\]
\[\deriv{f}(x) = \tolim{\delx}{0}\frac{f(x + \delx) - f(x)}{\delx} = \tolim{\delx}{0} \frac{c - c}{\delx} = 0 \eqper \qedhere\]
\end{proof}
\begin{example}
@@ -116,7 +116,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
为此，考虑
\[\tolim{\delx}{0}\frac{1}{\delx}\left[\frac{1}{g(x+\delx)}-\frac{1}{g(x)}\right] = \tolim{\delx}{0} -\frac{g(x+\delx)g(x)}{\delx} \cdot \frac{1}{g(x+\delx)g(x)} = - \frac{\deriv{g}(x)}{g^2(x)}\]
对于一般的$f$可由乘法求导公式得到
\[\deriv{\left(\frac{f}{g}\right)} = \deriv{\left(f \cdot \frac{1}{g}\right)} = \deriv{f}\frac{1}{g} + f \deriv{\left(\frac{1}{g}\right)} = \frac{\deriv{f}g-f\deriv{g}}{g^2}\eqper\]
\[\deriv{\left(\frac{f}{g}\right)} = \deriv{\left(f \cdot \frac{1}{g}\right)} = \deriv{f}\frac{1}{g} + f \deriv{\left(\frac{1}{g}\right)} = \frac{\deriv{f}g-f\deriv{g}}{g^2}\eqper \qedhere\]
\end{proof}
\begin{example}
@@ -173,7 +173,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
令$\Delta t \to 0$，则$\alpha (\Delta x) \to 0 \quad (\Delta x \to 0)$，式\eqref{链式法则式2}可以化为
\begin{align*}
\frac{\Delta y}{\Delta x} & = \deriv{f}(x) - \deriv{\varphi}(b)\\
& = \deriv{f}(\varphi(t)) \deriv{\varphi}(t)\eqper
& = \deriv{f}(\varphi(t)) \deriv{\varphi}(t)\eqper \qedhere
\end{align*}
\end{proof}
@@ -243,7 +243,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
\[y = \psi[\invertfunc{\varphi}(x)]\]
再利用复合函数和反函数微分法，得
\[\frac{\dif y}{\dif x} = \frac{\dif y}{\dif t} \cdot \frac{\dif t}{\dif x} = \frac{\frac{\dif y}{\dif t}}{\frac{\dif x}{\dif t}} = \frac{\deriv{\psi}(t)}{\deriv{\varphi}(t)}\]
\[\frac{\dif y}{\dif x} = \frac{\dif y}{\dif t} \cdot \frac{\dif t}{\dif x} = \frac{\frac{\dif y}{\dif t}}{\frac{\dif x}{\dif t}} = \frac{\deriv{\psi}(t)}{\deriv{\varphi}(t)} \eqper \qedhere\]
\end{proof}
\subsection{基本导数公式}
@@ -367,7 +367,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
\begin{proof}
构造函数$F(x) = f(x) - \frac{f(b) - f(a)}{b-a}(x-a)$，$F(x) \in C[a,b]$。注意到$F(b) f(b) - [f(b) - f(a)] = f(a) = F(a)$，应用Rolle定理，$\exists c \in (a,b)$使得$\deriv{F}(c) = 0$，即
\[\deriv{F}(c) = \deriv{f}(c) - \frac{f(b)-f(a)}{b-a} = 0\eqper\]
\[\deriv{F}(c) = \deriv{f}(c) - \frac{f(b)-f(a)}{b-a} = 0\eqper \qedhere\]
\end{proof}
\begin{remark}
@@ -464,7 +464,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
x \in (x_0, x_0 + \delta),\ \deriv{f}(x) \leq 0,\ \therefore f(x) \leq f(x_0)
\end{aligned}
\right\}
\text{极大值}
\text{极大值}\qedhere
\end{equation*}
\end{proof}
@@ -489,7 +489,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
x \in (x_0, x_0 + \delta),\ \deriv{f}(x) < 0,\ \therefore f(x) < f(x_0)
\end{aligned}
\right\}
\text{严格极大值}
\text{严格极大值}\qedhere
\end{equation*}
\end{proof}
@@ -518,7 +518,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
\begin{align*}
f(\lambda_1x_1 + \lambda_2x_2 + \lambda_3x_3) & \leq \lambda_1f(x_1) + (\lambda_2 + \lambda_3)f\left(\frac{\lambda_2 x_2 + \lambda_3 x_3}{\lambda_2 + \lambda_3}\right)\\
& \leq \lambda_1f(x_1) + (\lambda_2 + \lambda_3)\left(\frac{\lambda_2}{\lambda_2 + \lambda_3}f(x_2) + \frac{\lambda_3}{\lambda_2 + \lambda_3}f(x_3)\right)\\
& = \lambda_1 f(x_1) + \lambda_2 f(x_2) + \lambda_3 f(x_3)
& = \lambda_1 f(x_1) + \lambda_2 f(x_2) + \lambda_3 f(x_3) \eqper \qedhere
\end{align*}
\end{proof}
@@ -585,7 +585,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
再利用初等分式不等式
\[\frac{a_1}{b_1} \leq \frac{a_2}{b_2}\text{且}b_1, b_2 > 0,\ \text{则}\frac{a_1}{b_1} \leq \frac{a_1 + a_2}{b_1 + b_2} \leq \frac{a_2}{b_2}\]
由此即可得到
\[\frac{f(x) - f(x_1)}{x - x_1} \leq \frac{f(x_2) - f(x_1)}{x_2 - x_1} \leq \frac{f(x_2) - f(x)}{x_2 - x}\eqper\]
\[\frac{f(x) - f(x_1)}{x - x_1} \leq \frac{f(x_2) - f(x_1)}{x_2 - x_1} \leq \frac{f(x_2) - f(x)}{x_2 - x}\eqper \qedhere\]
\end{proof}
\begin{corollary}[二阶导数判别凸性]
@@ -604,7 +604,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
补充定义$f(a) = g(a) = 0$，从而使得$f,g$在$[a,b)$上连续。利用Cauchy中值定理，对$x \in (a,b)$，存在$a < \xi < x$使得
\[\frac{f(x)}{g(x)} = \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{\deriv{f}(\xi)}{\deriv{g}(\xi)}\]
那么当$x \to a^+$时，有$\xi \to a^+$，因此
\[\tolim{x}{a^+} \frac{f(x)}{g(x)} = \tolim{x}{a^+} \frac{\deriv{f}(\xi)}{\deriv{g}(\xi)} = \tolim{x}{a^+}\frac{\deriv{f}(x)}{\deriv{g}(x)}\eqper\]
\[\tolim{x}{a^+} \frac{f(x)}{g(x)} = \tolim{x}{a^+} \frac{\deriv{f}(\xi)}{\deriv{g}(\xi)} = \tolim{x}{a^+}\frac{\deriv{f}(x)}{\deriv{g}(x)}\eqper \qedhere\]
\end{proof}
\begin{remark}
@@ -624,7 +624,7 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
那么
\begin{align*}
\tolim{x}{+\infty} & = \tolim{t}{0^+}\frac{f \left(\dfrac{1}{t}\right)}{g \left(\dfrac{1}{t}\right)} = \tolim{t}{0^+}\frac{\deriv{f}\left(\dfrac{1}{t}\right)\left(-\dfrac{1}{t^2}\right)}{\deriv{g}\left(\dfrac{1}{t}\right)\left(-\dfrac{1}{t^2}\right)}\\
& = \tolim{t}{0^+}\frac{\deriv{f}\left(\dfrac{1}{t}\right)}{\deriv{g}\left(\dfrac{1}{t}\right)} = \tolim{x}{+\infty}\frac{\deriv{f}(x)}{\deriv{g}(x)}\eqper
& = \tolim{t}{0^+}\frac{\deriv{f}\left(\dfrac{1}{t}\right)}{\deriv{g}\left(\dfrac{1}{t}\right)} = \tolim{x}{+\infty}\frac{\deriv{f}(x)}{\deriv{g}(x)}\eqper \qedhere
\end{align*}
\end{proof}