# Quick LaTeX Tutorial

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$2!$

Should be fixed now. It was a bug in how the software handled exclamation marks -- I didn't realize IPB escapes them to their HTML entities.

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Pardon the interuption, but I was trying to use LaTeX in the astronomy and cosmology section and it didn't seem to work. I'm just testing a short script here;

$\frac{2\times\pi\times8.8\times10^8 m}{8834s}\eq\frac{625900 m}{s}$

[ math]\frac{2\times\pi\times8.8\times10^8 m}{8834s}\eq\frac{625900 m}{s}[/math ]

I inserted the spaces after the beginning and before the ending brackets so that the quoted section would appear as the original script I wrote.

Well...obviously I'm doing something wrong. The equation I'm trying to write is:

(2*pi*8.8*10^8 m)/(8834 s)=625900 m/s

Can anyone tell me what I'm doing wrong?

Chris

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Pardon the interuption, but I was trying to use LaTeX in the astronomy and cosmology section and it didn't seem to work. I'm just testing a short script here;

$\frac{2\times\pi\times8.8\times10^8 m}{8834s}\eq\frac{625900 m}{s}$

Try this:

$\frac{2\times\pi\times8.8\times10^8 m}{8834s} = 625900 m/s$

$\frac{2\times\pi\times8.8\times10^8 m}{8834s} = 625900 m/s$

or this:

$\frac{2\times\pi\times8.8\times10^8 m}{8834s} = 625900 m{s}^{-1}$

$\frac{2\times\pi\times8.8\times10^8 m}{8834s} = 625900 m{s}^{-1}$

Edited by murshid

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$\frac{2\times\pi\times8.8\times10^8\,m}{8834\,s}=\frac{625900\,m}{s}=625,900\,m/s=625.9\,km/s$

[ math]\frac{2\times\pi\times8.8\times10^8\,m}{8834\,s}=\frac{625900\,m}{s}=625,900\,m/s=625.9\,km/s[/math ]

It seems that what tripped me up is the use of "\eq" for the equals symbol instead of just inserting "=" in the script. Also, it's easier to read the equation if I use "\," before "m" and "s". I remember reading in one of the tutorial pages that spaces between characters are ignored.

Thanks for the help.

Chris

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You can also use \mbox{} for the m/s. Click on the equation to see the code used to make it:

$625,900 \, \mbox{m/s}$

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Please, help me install IPBlatex on IPBoard.

If I want to install IPBLatex, I should have any program on my computer?

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just testing

$A \bigcup B$

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just testing

$A \bigcup B$

You can test here: http://www.scienceforums.net/topic/58931-latex-testing/

Edited by StringJunky

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$e = mc^2$

Edited by DevilSolution

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How do I write matrices in latex, ie

a b

c d

?

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\left( \begin{array}{ccc}a & b & c \\d & e & f \\g & h & i \end{array} \right)

$\left( \begin{array}{ccc}a & b & c \\d & e & f \\g & h & i \end{array} \right)$

There is probably an easier way

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You can also try:

\begin{pmatrix}a & b & c \\d & e & f \\g & h & i \end{pmatrix}

$\begin{pmatrix}a & b & c \\d & e & f \\g & h & i \end{pmatrix}$

like this? /neq

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like this? /neq

no. You type \neq and then highlight it and select latex from the drop down lists accessed on the third button from the left of the top row. You then get

$\neq$

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Thank you so much. This guide was incredibly useful to me!

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You didn't include overbar!

\bar{x}

$\bar{x}$

But this was the first thing I couldn't find in the tutorial. nice job

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I was wondering if there is an easy answer about this. There are multiple versions of Latex, and I'm still a newbie to it. I was wondering if you guys knew which version is best to use and why? Also is there any fundamental differences between the different versions, and how much of the latex is standardized?

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$b_n= \frac{1}{T} \int^{T}_{0} f(t) \sin( \frac{2 \pi nt}{T})dt$

$x^2$

$xt=a0+\sum[a_k * \cos(\lambda_k * t) + b_k*\sin(\lambda_k*t)]$

$\lambda = 2 * \pi * \nu_k$

$\nu_k = k/q$

$a_0 = \frac{1}{T} \int^{T}_{0} f(t) dt$

$\frac{cov}{\sum^{n}_{i = 1}(x_i - \bar{x})^2 }$

$\sum^{n}_{i = 1}$

Edited by DevilSolution

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$70*\tn$



Edited by DevilSolution

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${\bar \lambda} = \frac {\lambda}{2 \pi} = \frac{\hbar}{m c}$

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It works

Edited by renerpho

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Notice that your sum is of the form $\sum_{n=1}^{m}\frac{P(n)}{2^n}$ where $P$ is a polynomial. You start your proof with an educated guess, that the sum will be of similar form, namely $\sum_{n=1}^{m}\frac{n^3}{2^n}\stackrel{?}{=}\frac{{Q{_1}(m)}2^m+Q{_2}(m)}{2^m}$ where $Q{_1}$ and $Q{_2}$ are themselves polynomials (you include a polynomial term multiplied by $2^m$ to increase your chances of success). You can be quite confident that $Q{_2}$ will be of at least same degree as $P$. So, your first attempt is the simplest possible, where $Q{_1}$ has degree 0 (turning it into a constant) and $Q{_2}$ has degree 3.

This leads you to $\sum_{n=1}^{m}\frac{n^3}{2^n}\stackrel{?}{=}\frac{{a}2^m+{b}m^3+{c}m^2+{d}m+e}{2^m}$ for some real numbers $a,b,c,d,e$.

Evaluate at $m=1,\cdots,5$ and you get a system of 5 linear equations in 5 variables. That means that IF your guess was correct then this will lead you to the only possible solution. That's the one presented above, and once you found it you can proof it by induction.

Edited by renerpho

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testing

$s^{/frac{1}{2}} = 0$

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testing

$s^{/frac{1}{2}} = 0$

your slash before the frac is the wrong way around

$s^{\frac{1}{2}}$

click on mine to see difference

Oh I see.

thanks.

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