Difference between revisions of "Sandbox"

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<syntaxhighlight  lang='python' line highlight='2'>
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{{LaTeX Shortcuts}}
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And a one
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$$\newcommand{\blah}{\alpha\beta}$$
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and a two
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$$\newcommand{\blaha}{\alpha\beta}$$
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and a three
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$$\newcommand{\blahb}{\alpha\beta}$$
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</syntaxhighlight>
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$$\newcommand{\blahc}{\alpha\beta}$$
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$$\newcommand{\blahd}{\alpha\beta}$$
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<syntaxhighlight  lang='python' highlight='2'>
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There is $$\frac{g \beta \Delta T H^3}{\alpha \nu}$$ and then there is
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And a one
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\[ \frac{g \beta \Delta T H^3}{\alpha \nu} \]
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and a two
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$$
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and a three
 
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</syntaxhighlight>
 
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<syntaxhighlight lang="matlab">
 
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clear
 
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format short e
 
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plot(x, y, 'k--')
 
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</syntaxhighlight>
 
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alpha: <math>1+1</math>
 
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$\alpha$
 
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$$\alpha$$
 
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<center><math>
 
 
\begin{align}
 
\begin{align}
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H&=\frac{j\omega+8}{(j\omega)^2+4*j\omega+13}\\
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\frac{g \beta \Delta T H^3}{\alpha \nu}
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H&=\frac{s+8}{(s+2)^2+(3)^2}=\frac{A(s+2)+B(3)}{(s+2)^2+(3)^2}
 
 
\end{align}
 
\end{align}
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</math></center>
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$$
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From the <math>s</math> in the numerator, you can see that <math>A</math> is 1.  That means the numerator so far is <math>1(s+2)</math>  Therefore, there is already a constant 2 up top.  To get the total of 8, then <math>B</math> needs to be 2.  That is:
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\[
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<center><math>
 
 
\begin{align}
 
\begin{align}
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H&=\frac{s+8}{(s+2)^2+(3)^2}=\frac{A(s+2)+B(3)}{(s+2)^2+(3)^2}=\frac{1(s+2)+2(3)}{(s+2)^2+(3)^2}
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\frac{g \beta \Delta T H^3}{\alpha \nu}
 
\end{align}
 
\end{align}
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</math></center>
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\]
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meaning
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<center><math>
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$$\blah$$
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h(t):=e^{-2t}\left(1*\cos(3t)+2*\sin(3t)\right)~u(t)
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</math></center>
 
  
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It is now possible to do inline math such as $$c=\sqrt{a^2+b^2}$$ with double dollar signs around the math.
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<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">

Latest revision as of 16:12, 11 February 2024

$$\newcommand{E}[2]{#1_{\mathrm{#2}}}$$ $$\newcommand{\blah}{\alpha\beta}$$ $$\newcommand{\blaha}{\alpha\beta}$$ $$\newcommand{\blahb}{\alpha\beta}$$ $$\newcommand{\blahc}{\alpha\beta}$$ $$\newcommand{\blahd}{\alpha\beta}$$ There is $$\frac{g \beta \Delta T H^3}{\alpha \nu}$$ and then there is \[ \frac{g \beta \Delta T H^3}{\alpha \nu} \] $$ \begin{align} \frac{g \beta \Delta T H^3}{\alpha \nu} \end{align} $$ \[ \begin{align} \frac{g \beta \Delta T H^3}{\alpha \nu} \end{align} \]

$$\blah$$


<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">

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