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qedhere是个好东西

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unlockable
2022-11-07 11:08:25 +08:00
parent 785f77973e
commit 58f36b9bb1
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12
04微分与Taylor定理.tex
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@@ -123,7 +123,7 @@
& = \cdots = \tolim{\Delta x}{0} \frac{f^{(n-1)}(x_0 + \Delta x) - P_n^{(n-1)}(\Delta x)}{n(n-1)\cdots 2\cdot \Delta x}\\
& = \tolim{\Delta x}{0} \frac{f^{(n-1)}(x_0 + \Delta x) - f^{(n-1)}(x_0) - f^{(n)}(x_0)\Delta x}{n!\Delta x}\\
& = \frac{1}{n!} \tolim{\Delta x}{0} \left[\frac{f^{(n-1)}(x_0 + \Delta x) - f^{(n-1)}(x_0)}{\Delta x} - f^{(n)}(x_0) \right]\\
& = 0
& = 0 \eqper \qedhere
\end{align*}
\end{proof}
@@ -133,7 +133,7 @@
\begin{proof}[解]
$f^{(k)}(x) = e^x$，$f^{(k)}(0) = 1$，$k = 1, 2, 3 \cdots$。因此
\[e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} + o(x^n)\eqper\]
\[e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} + o(x^n)\eqper \qedhere\]
\end{proof}
\begin{example}
@@ -151,7 +151,7 @@
,\quad m = 0, 1, 2\cdots
\end{equation*}
所以
\[\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots + (-1)^m \frac{x^{2m+1}}{(2m+1)!} + o(x^{2m+1})\eqper\]
\[\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots + (-1)^m \frac{x^{2m+1}}{(2m+1)!} + o(x^{2m+1})\eqper \qedhere\]
\end{proof}
\begin{example}
@@ -169,7 +169,7 @@
,\quad m = 0, 1, 2\cdots
\end{equation*}
所以
\[\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots + (-1)^m\frac{x^{2m}}{(2m)!} + o(x^{2m})\eqper\]
\[\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots + (-1)^m\frac{x^{2m}}{(2m)!} + o(x^{2m})\eqper \qedhere\]
\end{proof}
\begin{example}
@@ -196,7 +196,7 @@
m = 0, 1, 2, \cdots
\end{equation*}
因此
\[\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots + (-1)^m \frac{x^{2m+1}}{2m + 1} + o(x^{2m+1})\eqper\]
\[\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots + (-1)^m \frac{x^{2m+1}}{2m + 1} + o(x^{2m+1})\eqper \qedhere\]
\end{proof}
\begin{example}
@@ -205,7 +205,7 @@
\begin{proof}
$f^{(k)}(0) = \alpha (\alpha - 1)\cdots(\alpha - k + 1),\quad k = 1, 2, \cdots$，因此
\[(1 + x)^\alpha = 1 + \alpha x + \frac{\alpha (\alpha - 1)}{2!}x^2 + \cdots + \frac{\alpha (\alpha - 1) \cdots (\alpha - n + 1)}{n!}x^n + o(x^n)\eqper\]
\[(1 + x)^\alpha = 1 + \alpha x + \frac{\alpha (\alpha - 1)}{2!}x^2 + \cdots + \frac{\alpha (\alpha - 1) \cdots (\alpha - n + 1)}{n!}x^n + o(x^n)\eqper \qedhere\]
\end{proof}
\begin{remark}