Tutorials too long. Quick guide?
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asked Aug 29, 2021 at 19:15
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2 Answers
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0
$x$
$x$
$$ x + y^{2k} $$
$$
x + y^{2k}
$$
Inline $x + y$ with text
Inline $x + y$ with text
$$ a \leq \frac{b}{c} $$
$$
a \leq \frac{b}{c}
$$
$$ y[n] = \sum_{k=0}^{m} b_k $$
$$
y[n] = \sum_{k=0}^{m} b_k
$$
$$ \mathbf M = \begin{pmatrix} 1 & 2 & \pi \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} $$
$$
\mathbf M =
\begin{pmatrix}
1 & 2 & \pi \\
4 & 5 & 6 \\
7 & 8 & 9
\end{pmatrix}
$$
$$ \int_{-\infty}^{\infty} \hat\psi_{a,b}(\omega) d\omega, \omega \in \mathbb{R} $$
$$
\int_{-\infty}^{\infty} \hat\psi_{a,b}(\omega) d\omega, \omega \in \mathbb{R}
$$
$$ f(n) = \begin{cases} n/2, & \text{if $n$ is even} \\ \sqrt{3n+1}, & \text{if $n$ is odd} \end{cases} $$
$$
f(n) =
\begin{cases}
n/2, & \text{if $n$ is even} \\
\sqrt{3n+1}, & \text{if $n$ is odd}
\end{cases}
$$
\begin{align} 5 &= 0.5^{-1} + 3 \\ &= 2 + 3 \\ &= \boxed{5} \tag{1}\label{1} \\ \end{align}
\begin{align}
5 &= 0.5^{-1} + 3 \\
&= 2 + 3 \\
&= \boxed{5} \tag{1}\label{1} \\
\end{align}
And in $\eqref{1}$ we see that the result is five.
And in $\eqref{1}$ we see that the result is five.
answered Aug 29, 2021 at 19:15
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1
Advanced
MathJax reference -- $$ omitted below
$$ \mathbf I = \begin{pmatrix} 1 & 0 & \cdots & 0\\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{pmatrix} $$
\mathbf I =
\begin{pmatrix}
1 & 0 & \cdots & 0\\
0 & 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 1
\end{pmatrix}
$$ \cos(\omega_0 t) \Leftrightarrow \pi \delta(|w| - \omega_0) $$
\cos(\omega_0 t) \Leftrightarrow \pi \delta(|w| - \omega_0)
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$\begingroup$ Feel free to add stuff here, I rather keep the main answer simple $\endgroup$ Sep 5, 2021 at 9:53