unlockable/MathematicalAnalysis
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

qedhere是个好东西

Browse Source
This commit is contained in:
unlockable
2022-11-07 11:08:25 +08:00
parent 785f77973e
commit 58f36b9bb1
4 changed files with 40 additions and 42 deletions
Download Patch File Download Diff File Expand all files Collapse all files

6
02函数及其连续性.tex
View File

@@ -268,7 +268,7 @@
所以
\[\lim \limits_{x \to 0} \frac{\sin x}{x} = 1\]
同时也得到
\[\lim \limits_{x \to 0} \cos x = 1 \eqper\]
\[\lim \limits_{x \to 0} \cos x = 1 \eqper \qedhere\]
\end{proof}
\begin{proposition}[单调收敛原理]
@@ -364,7 +364,7 @@
\begin{proof}
\begin{equation*}
\toxzero u(x)^{v(x)} = \toxzero e^{v(x) \ln u(x)} = e^{\toxzero \left(v(x) \ln u(x)\right)} = e^{\toxzero v(x) \cdot \toxzero \ln u(x)} = e^{b \ln a} = a^b \eqper
\toxzero u(x)^{v(x)} = \toxzero e^{v(x) \ln u(x)} = e^{\toxzero \left(v(x) \ln u(x)\right)} = e^{\toxzero v(x) \cdot \toxzero \ln u(x)} = e^{b \ln a} = a^b \eqper \qedhere
\end{equation*}
\end{proof}
@@ -824,7 +824,7 @@ $x = 0$是$f$的第二类间断点。
令$s = x$得
\[\vert B_n(f)(x) - f(x) \vert \leq \frac{\varepsilon}{2}+\frac{2M}{n\delta^2}\left(x-x^2\right) \leq \frac{\varepsilon}{2} + \frac{M}{2n\delta^2}\eqco \forall x \in [0,1]\]
任意取定$n > \dfrac{M}{\delta^2 \varepsilon}$，有
\[\vert B_n(f)(x) - f(x) \vert = \left|\sum_{k=0}^n f\left(\frac{k}{n}\right)C_n^k x^k (1-x)^{n-k} - f(x)\right| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \eqco x \in [0,1]\]
\[\vert B_n(f)(x) - f(x) \vert = \left|\sum_{k=0}^n f\left(\frac{k}{n}\right)C_n^k x^k (1-x)^{n-k} - f(x)\right| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \eqco x \in [0,1]\eqper \qedhere\]
\end{proof}
\begin{remark}