This weeks video choice was easy... Very easy... From a galaxy far far away...
Friday, 13 October 2017
Tuesday, 3 October 2017
MathJax on Blogger Cheatsheet
1. Placement
$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$
will generate the formula inline \(\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}\) of a paragraph, where
$$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$$
will render the formula as a seperate image $$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$$ and not inline to the paragraph.2. Greek letters
For Greek letters, use
$\alpha$
\(\alpha\)$\beta$
\(\beta\)$\delta$
\(\delta\) or$\Delta$
\(\Delta\)$\gamma$
\(\gamma\) or$\Gamma$
\(\Gamma\)$\ldots$
\(\ldots\)$\omega$
\(\omega\) or$\Omega$
\(\Omega\)
3. Superscripts and subscripts
For superscripts and subscripts, use
^
and _
.
$x_i^2$
\(x_i^2\)$\log_2x$
\(\log_2x\)
4. Grouping
Superscripts, subscripts, and other operations apply only to the next “group”. A “group” is either a single symbol, or any formula surrounded by curly braces {…}.
If you do
$10^10$
, you will get a surprise: \(10^10\). But 10^{10}
gives what you probably wanted: \(10^{1}\).Use curly braces to delimit a formula to which a superscript or subscript applies:
$x^5^6$
is an error; ${x^y}^z$
is \({x^y}^z\), and $x^{y^z}$
is \(x^{y^z}\).Observe the difference between
$x_i^2$
\(x_i^2\) and $x_{i^2}$
\(x_{i^2}\).5. Parentheses
Ordinary symbols
$()[]$
make parentheses and brackets \((2+3)[4+4](2+3)[4+4]\).Use
$\{$
and $\}$
for curly braces \(\{\ldots\}\) and use $\($
and $\)$
for round braces \((\ldots)\).These do not scale with the formula in between, so if you write
$(\frac{\sqrt x}{y^3})$
the parentheses will be too small: \(\frac{\sqrt x}{y^3}\).Using
$\left$
and $\right$
will make the sizes adjust automatically to the formula they enclose: $\left(\frac{\sqrt x}{y^3}\right)$
is \(\left(\frac{\sqrt x}{y^3}\right)\).6. Sums and integrals
$\sum$
and $\int
the subscript is the lower limit and the superscript is the upper limit, so for example $\sum_1^n$
renders as \(\sum_1^n\).Remember that
{…}
if the limits are more than a single symbol. For example, $\sum_{i=0}^\infty i^2$
renders as \(\sum_{i=0}^\infty i^2\).Similarly,
$\prod$
\(\prod\), $\int$
\(\int\), $\bigcup$
\(\bigcup\), $\bigcap$
\(\bigcup\) and/or \iint
\(\iint\).7. Fractions
There are two approaches
$\frac ab$
applies to the next two groups renders as \(\frac ab\); and more complicated numerators and denominators use
$/{…/}$
$\frac{a+1}{b+1}$
renders as \(\frac{a+1}{b+1}\).
If the numerator and denominator are complicated, you may prefer \over, which splits up the group that it is in, eg. ${a+1\over b+1}$
renders as \({a+1\over b+1}\).8. Radical signs
Use
$\sqrt$
, which adjusts to the size of its argument, eg. $\sqrt{x^3}$
renders as \(\sqrt{x^3}\) and $\sqrt[3]{\frac xy}$
renders as \(\sqrt[3]{\frac xy}\).For complicated expressions, consider using
${...}^{1/2}$
instead.9. Special functions
Such as
$\lim$
, $\sin$
, $\max$
, $\ln$
, etc. are normally set in roman font instead of italic font. $\sin x$
renders as \(\sin x\), and not $/sin x$
renders as \(sin x\).Use subscripts to attach a notation to
$\lim$
, eg. $\lim_{x\to 0}$
renders as \(\lim_{x\to 0}\).10. Some special symbols and notations
$\lt$
\(\lt\) and$\not\lt$
\(\not\lt\)$\gt$
\(\gt\) and$\not\gt$
\(\not\gt\)$\le$
\(\le\) and$\not\le$
\(\not\le\)$\ge$
\(\ge\) and$\not\ge$
\(\not\ge\)$\times$
\(\times\) and$\div$
\(\div\)$\pm$
\(\pm\) and$\mp$
\(\mp\)$\ell
\(\ell\)$\cdot$
\(\cdot\) and$\cup$
\(\cup\) and$\cap$
\(\cap\)$\setminus$
\(\setminus\)$\subset$
\(\subset\) and$\subseteq$
\(\subseteq\) and$\subsetneq$
\(\subsetneq\) and$\supset$
\(\supset\)$\in$
\(\in\) and$\notin$
\(\notin\)$\emptyset$
\(\emptyset\) and$\varnothing$
\(\varnothing\)${n+1 \choose 2k}$
or$${n+1 \choose 2k}$
or$\binom{n+1}{2k}$
\(n+1 \choose 2k\)$\to$
\(\to\) or$\rightarrow$
\(\rightarrow\) or$\leftarrow$
\(\leftarrow\) or$\Rightarrow$
\(\Rightarrow\) or$\Leftarrow$
\(\Leftarrow\) or$\mapsto$
\(\mapsto\)$\land$
\(\land\) or$\lor$
\(\lor\) or$\lnot$
\(\lnot\) or$\forall$
\(\forall\) or$\exists$
\(\exists\) or$\top$
\(\top\) or$\bot$
\(\bot\) or$\vdash$
\(\vdash\) or$\vDash$
\(\vDash\)$\star$
\(\star\) or$\ast$
\(\ast\) or$\oplus$
\(\oplus\) or$\circ$
\(\circ\) or$\bullet$
\(\bullet\)$\approx$
\(\approx\) or$\sim$
\(\sim\) or$\simeq$
\(\simeq\) or$\cong$
\(\cong\) or$\equiv$
\(equiv\) or$\prec$
\(\prec\) or$\lhd$
\(\lhd\)$\infty$
\(\infty\) and$\aleph_0$
\(\aleph_0\) or$\nabla$
\(\nabla\) and$\partial$
\(\partial\) or$\Im$
\(\Im\) or$\Re$
\(\Re\)$\pmod$
for modular equivalence eg.$a\equiv b\pmod n$$
would render as \(a\equiv b\pmod n\)$\ldots
is the dots in$a1,a2,\ldots,an$
renders as \(a1,a2,\ldots,an\) and$\cdots$
is the dots in$a1+a2+⋯+ana1+a2+⋯+an$
renders as \(a1+a2+\cdots+ana1+a2+\cdots+an\)$\epsilon$
renders as \(\epsilon\) and$\varepsilon$
renders as \(\varepsilon\)$\phi$
renders as \(\phi\) and$\varphi$
renders as \(\varphi\)
11. Spaces
$a_b$
and $a____b$
are both \(ab\).
$\,$
renders as \(a\,b\)$\;$
renders as \(a\;b\)$\quad$
renders as \(a \quad b\)$\qquad$
renders as \(a \qquad b\)
$sum_(i=1)^n i^3=((n(n+1))/2)^2$
renders as
\(sum_(i=1)^n i^3=((n(n+1))/2)^2\)$\begin{align}
\sqrt{37}
& = \sqrt{\frac{73^21}{12^2}} \\
& = \sqrt{\frac{73^2}{12^2}\cdot\frac{73^21}{73^2}} \\
& = \sqrt{\frac{73^2}{12^2}}\sqrt{\frac{73^21}{73^2}} \\
& = \frac{73}{12}\sqrt{1  \frac{1}{73^2}} \\
& \approx \frac{73}{12}\left(1  \frac{1}{2\cdot73^2}\right)
\end{align}$
renders as
\(\begin{align} \sqrt{37} & = \sqrt{\frac{73^21}{12^2}} \\ & = \sqrt{\frac{73^2}{12^2}\cdot\frac{73^21}{73^2}} \\ & = \sqrt{\frac{73^2}{12^2}}\sqrt{\frac{73^21}{73^2}} \\ & = \frac{73}{12}\sqrt{1  \frac{1}{73^2}} \\ & \approx \frac{73}{12}\left(1  \frac{1}{2\cdot73^2}\right) \end{align} \)$P(Z\le z) = \Phi(z) = \int_{\infty}^z \frac{1}{\sqrt{2\pi}} e^{w^2/2}\,dw$
renders as \(P(Z\le z) = \Phi(z) = \int_{\infty}^z \frac{1}{\sqrt{2\pi}} e^{w^2/2}\,dw\)$\begin{align} \nabla \times \vec{\mathbf{B}} \, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\ \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\ \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\ \nabla \cdot \vec{\mathbf{B}} & = 0 \end{align}$
renders as \(\begin{align} \nabla \times \vec{\mathbf{B}} \, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\ \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\ \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\ \nabla \cdot \vec{\mathbf{B}} & = 0 \end{align}\)
 MathJAX  https://www.mathjax.org
 Calculatorium.com  http://www.calculatorium.com/mathjaxquickstarttutorial/
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