[英]How to write piecewise function in jupyter notebook markdown?
I'm trying to write a piecewise function in markdown on jupyter and jupyter is being spooky. 我正在尝试在jupyter的markdown中编写分段功能,而jupyter令人毛骨悚然。 Out of three functions, it only displays one properly (the $n_o$ one), even though they are all written in the same manner. 在这三个函数中,即使它们都是以相同的方式编写的,它也只能正确显示一个($ n_o $一个)。
Is there a more elegant way to write this up? 有没有更优雅的方式来写这个?
$$ N(a)= \left\{
\begin{array}{ll}
n_o & A>A_{krit} \\
n_o+2 & A=A_{krit} \\
n_o+4 & A<A_{krit} \\
\end{array}
\right, $$
gdje je: $$A=\frac{1}{|a|},$$
$$ A_{krit}(a)= \left\{
\begin{array}{ll}
\frac{\sin(2\pi*floor(\frac{|a|}{2*\pi})+\frac{\pi}{2})}{2\pi*floor(\frac{|a|}{2*\pi})+\frac{\pi}{2}} & a>0 \\
\frac{\sin(2\pi*floor(\frac{|a|}{2*\pi})+\frac{3\pi}{2})}{2\pi*floor(\frac{|a|}{2*\pi})+\frac{3\pi}{2}} & a<0\\
\end{array}
\right, $$ i
$$ n_o(a)= \left\{
\begin{array}{ll}
|floor(\frac{|a|}{2*\pi})-1| & a>2\pi \\
|floor(\frac{|a|}{2*\pi})-1|-2 & 0<a<2\pi \land A\leq A_{krit} \\
floor(\frac{|a|}{2*\pi})+1 & a<0 \\
\end{array}
\right. $$
It seems there is spacing issue in your latex code. 您的乳胶代码中似乎存在间距问题。 I just modified little bit, it works: 我只是修改了一点,它的工作原理是:
$$ N(a)= \left\{
\begin{array}{ll}
n_o & A>A_{krit} \\
n_o+2 & A=A_{krit} \\
n_o+4 & A<A_{krit} \\
\end{array}
\right. $$
gdje je: $$A=\frac{1}{|a|},$$
$$ A_{krit}(a)= \left\{
\begin{array}{ll}
\frac{\sin(2\pi*floor(\frac{|a|}{2*\pi})+\frac{\pi}{2})}{2\pi*floor(\frac{|a|}{2*\pi})+\frac{\pi}{2}} & a>0 \\
\frac{\sin(2\pi*floor(\frac{|a|}{2*\pi})+\frac{3\pi}{2})}{2\pi*floor(\frac{|a|}{2*\pi})+\frac{3\pi}{2}} & a<0\\
\end{array}
\right. $$ i
$$ n_o(a)= \left\{
\begin{array}{ll}
|floor(\frac{|a|}{2*\pi})-1| & a>2\pi \\
|floor(\frac{|a|}{2*\pi})-1|-2 & 0<a<2\pi \land A\leq A_{krit} \\
floor(\frac{|a|}{2*\pi})+1 & a<0 \\
\end{array}
\right. $$
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