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6 changes: 3 additions & 3 deletions 404.html
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Expand Down Expand Up @@ -948,7 +948,7 @@ <h1>The End of Internet (404 - Not found)</h1>
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8 changes: 4 additions & 4 deletions course/artificial-intelligence/index.html
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Expand Down Expand Up @@ -999,7 +999,7 @@ <h2 id="Bias-vs-Variance">Bias vs. Variance<a class="headerlink" href="#Bias-vs-
&amp;= \overline{(\hat y - \overline{\hat y})^2}
+ \overline{\qty(\overline{\hat y} - y)^2}
+ 2 \overline{\qty(\hat y - \overline{\hat y}) (\overline{\hat y} - y)} \\
&amp;=
&amp;=
\end{split}
\]</div>
<p>???</p>
Expand Down Expand Up @@ -1113,7 +1113,7 @@ <h2 id="__comments">评论</h2>
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26 changes: 13 additions & 13 deletions course/calculus-1/index.html
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Expand Down Expand Up @@ -1692,14 +1692,14 @@ <h4 id="双曲函数与三角函数的对应">双曲函数与三角函数的对
<p>2021年4月23日:</p>
<p>事实上双曲函数与三角函数的关系和反关系一样。例如</p>
<div class="arithmatex">\[
\sinh x
\sinh x
= \i \sin\frac x\i
= \i \sin\frac{\i x}{\i^2}
= \i \sin(-\i x)
= -\i \sin(\i x)
= \frac{\sin(\i x)}{\i}
\]</div>
<p>这与 <span class="arithmatex">\(\sin x =\frac{\sinh(\i x)}{\i}\)</span> 完全一致。 </p>
<p>这与 <span class="arithmatex">\(\sin x =\frac{\sinh(\i x)}{\i}\)</span> 完全一致。</p>
</blockquote>
<h4 id="一些等式">一些等式<a class="headerlink" href="#一些等式" title="Permanent link">&para;</a></h4>
<div class="arithmatex">\[
Expand Down Expand Up @@ -1735,8 +1735,8 @@ <h4 id="定义与性质">定义与性质<a class="headerlink" href="#定义与
<p>复合函数 <span class="arithmatex">\(x_1 \mathop{\rightarrow}^{f_{12}} x_2 \mathop{\rightarrow}^{f_{23}} x_3 \\\)</span> 的极限:</p>
<div class="arithmatex">\[
\lim_{x_1\to L_1} f_{12}(x_1) = L_2 \land
f_{12}(x_1)\neq L_2 \land
\lim_{x_2\to L_2} f_{23}(x_2) = L_3
f_{12}(x_1)\neq L_2 \land
\lim_{x_2\to L_2} f_{23}(x_2) = L_3
\quad\Rightarrow\quad
\lim_{x_1\to L_1} f_{23}\circ f_{12} \circ x_1 = L_3
\]</div>
Expand Down Expand Up @@ -1866,7 +1866,7 @@ <h4 id="1infty-0infty"><span class="arithmatex">\(1^\infty,\ 0^\infty\)</span><a
<div class="arithmatex">\[
\displaylines{
\alpha \to 0,\ \beta \to \infty,\ \alpha\beta \to L \\
\Rightarrow (1+\alpha)^\beta
\Rightarrow (1+\alpha)^\beta
= (1+\alpha)^{\frac{1}{\alpha} \cdot \alpha\beta}
= \left( (1+\alpha)^{\frac 1\alpha}\right) ^ {\alpha\beta}
\to \e^L
Expand Down Expand Up @@ -1926,7 +1926,7 @@ <h4 id="带Peano余项的Taylor展开">带Peano余项的Taylor展开<a class="he
\begin{split}
\lim_{x\to0} \frac{\sqrt{1+x}+\sqrt{1-x}-2}{x^2}
&amp;= \lim_{x\to0} \frac{\left(1 +\frac{x}2 -\frac{x^2}{8} + o(x^2) \right) +
\left(1 +\frac{-x}2 -\frac{(-x)^2}{8} + o(x^2) \right) -
\left(1 +\frac{-x}2 -\frac{(-x)^2}{8} + o(x^2) \right) -
2}{x^2} \\
&amp;= \lim_{x\to0} \frac{-\frac{x^2}4 + o(x^2)}{x^2}
= -\frac14
Expand Down Expand Up @@ -1992,7 +1992,7 @@ <h4 id="目的">目的<a class="headerlink" href="#目的" title="Permanent link
\begin{split}
\int \dfrac{c_1x + c_0}{(x^2 - 2\alpha x + \beta)^\mu} \d x
&amp;= \int \dfrac{c_1(x-\alpha) + c_1\alpha + c_0}{((x-\alpha)^2 + \beta - \alpha^2)^\mu} \d x \\
&amp;= \frac{c_1}{2(\beta-\alpha^2)^{\mu-1}} \cdot \int\frac{\d \frac{(x-\alpha)^2}{\beta-\alpha^2}} {\left(\frac{(x-\alpha)^2}{\beta-\alpha^2} + 1 \right)^\mu} +
&amp;= \frac{c_1}{2(\beta-\alpha^2)^{\mu-1}} \cdot \int\frac{\d \frac{(x-\alpha)^2}{\beta-\alpha^2}} {\left(\frac{(x-\alpha)^2}{\beta-\alpha^2} + 1 \right)^\mu} +
\frac{c_1\alpha + c_0}{(\beta-\alpha^2)^{\mu-\frac12}} \cdot \int\frac{\d \frac{x-\alpha}{\sqrt{\beta-\alpha^2}}} {\left(\frac{(x-\alpha)^2}{\beta-\alpha^2} + 1 \right)^\mu} \\
&amp;\rightarrow
\int \frac{\d u}{(u+1)^\mu} + \int \frac{\d x}{(x^2+1)^\mu}
Expand Down Expand Up @@ -2055,7 +2055,7 @@ <h4 id="特殊">特殊<a class="headerlink" href="#特殊" title="Permanent link
\begin{split}
\int \frac{c_1x + c_0}{x^2 - 2\alpha x + \beta} \d x
&amp;= \frac{c_1}2 \ln(x^2 - 2\alpha x + \beta) + \int \frac{c_1 \alpha + c_0}{x^2 - 2\alpha x + \beta}\d x \\
&amp;= \frac{c_1}2 \ln(x^2 - 2\alpha x + \beta) +
&amp;= \frac{c_1}2 \ln(x^2 - 2\alpha x + \beta) +
\frac{c_1\alpha+c_0}{\sqrt{\beta-\alpha^2}}\cdot \arctan \frac{x-\alpha}{\sqrt{\beta - \alpha^2}} +
\const
\end{split}
Expand All @@ -2075,8 +2075,8 @@ <h3 id="三角有理式">三角有理式<a class="headerlink" href="#三角有
\displaylines{
\int \tan^2 x\d x = \int (\sec^2 x - 1)\d x = \tan(x) - x + \const \\
\begin{split}
\int \frac{x\d x}{1+\cos x}
&amp;= \int \frac{x}{2} \sec^2\frac{x}{2} \d x
\int \frac{x\d x}{1+\cos x}
&amp;= \int \frac{x}{2} \sec^2\frac{x}{2} \d x
= \int x\d \tan\frac{x}{2} \\
&amp;= x \tan\frac{x}{2} - \int \tan\frac{x}{2}\d x \\
&amp;= x \tan\frac{x}{2} + 2\ln \cos\frac{x}{2} + \const
Expand Down Expand Up @@ -2314,7 +2314,7 @@ <h2 id="__comments">评论</h2>
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8 changes: 4 additions & 4 deletions course/calculus-2/index.html
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Expand Up @@ -19,7 +19,7 @@
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Expand Down Expand Up @@ -1594,7 +1594,7 @@ <h3 id="积分元素">积分元素<a class="headerlink" href="#积分元素" tit
\d x &amp; \d x\d y &amp; \d x\d y\d z \\
&amp; \rho\d\theta\d\rho &amp; \rho\d\theta\d\rho\d z \\
&amp; &amp; r\sin\varphi\d\theta\ r\d\varphi \d r \\
&amp; &amp; = \d\theta \sin\varphi\d\varphi\ r^2\d r
&amp; &amp; = \d\theta \sin\varphi\d\varphi\ r^2\d r
\end{array}
\]</div>
<h3 id="三角函数单项式的定积分">三角函数单项式的定积分<a class="headerlink" href="#三角函数单项式的定积分" title="Permanent link">&para;</a></h3>
Expand Down Expand Up @@ -2002,7 +2002,7 @@ <h2 id="__comments">评论</h2>
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Expand Up @@ -19,7 +19,7 @@
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Expand Down Expand Up @@ -1542,7 +1542,7 @@ <h3 id="其它方法">其它方法<a class="headerlink" href="#其它方法" tit
<div class="arithmatex">\[
\begin{split}
&amp;\oint\limits_C \frac{z^{15} \dd{z}} {\qty(z^2-1)^2 \qty(z^4+2)} \\
&amp;= \oint\limits_{2C} \frac{u^{7} \dd{\frac u2}} {\qty(u-1)^2 \qty(u^2+2)}
&amp;= \oint\limits_{2C} \frac{u^{7} \dd{\frac u2}} {\qty(u-1)^2 \qty(u^2+2)}
\qquad (u=z^2) \\
&amp;= \oint\limits_{C} \frac{u^{7} \dd{u}} {\qty(u-1)^2 \qty(u^2+2)} \\
&amp;= - 2\pi i\ \Res(\frac{u^7} {\qty(u-1)^2 \qty(u^2+2)}, \infty) \\
Expand Down Expand Up @@ -1746,7 +1746,7 @@ <h4 id="线性非齐次常系数微分方程">线性非齐次常系数微分方
<p>现在一点儿一点儿积分回去。</p>
<div class="arithmatex">\[
\begin{split}
e^{2ikx} u'
e^{2ikx} u'
&amp;= 0+ \int\limits_0^x \eval{\qty(e^{2ikx}u')'}_\xi \dd{\xi} \\
&amp;= \int\limits_0^x e^{ik\xi} \eval{f}_\xi \dd{\xi}. \\
\end{split}
Expand Down Expand Up @@ -1831,7 +1831,7 @@ <h3 id="调和方程的格林函数法">调和方程的格林函数法<a class="
\]</div>
<p>移项,</p>
<div class="arithmatex">\[
\eval{u}_{\vb* x}
\eval{u}_{\vb* x}
= \iiint \eval{G}_{\vb* x, \vb* x_0} \eval{F}_{\vb* x_0} \dd{V_0}
+ \oiint \eval{(u \grad_0 G - G \grad_0 u)}_{\vb* x_0} \vdot \dd{\vb* S_0}.
\]</div>
Expand Down Expand Up @@ -2172,7 +2172,7 @@ <h2 id="__comments">评论</h2>
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6 changes: 3 additions & 3 deletions course/cpp-and-data-structure/index.html
Original file line number Diff line number Diff line change
Expand Up @@ -19,7 +19,7 @@
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