Cap'n Refsmmat 1351 Posted November 8, 2010 [math]2![/math] 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. 0 Share this post Link to post Share on other sites

csmyth3025 27 Posted July 12, 2011 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; [math]\frac{2\times\pi\times8.8\times10^8 m}{8834s}\eq\frac{625900 m}{s}[/math] [ 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 0 Share this post Link to post Share on other sites

murshid 0 Posted July 12, 2011 (edited) 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; [math]\frac{2\times\pi\times8.8\times10^8 m}{8834s}\eq\frac{625900 m}{s}[/math] Try this: [math]\frac{2\times\pi\times8.8\times10^8 m}{8834s} = 625900 m/s[/math] [math]\frac{2\times\pi\times8.8\times10^8 m}{8834s} = 625900 m/s[/math] or this: [math]\frac{2\times\pi\times8.8\times10^8 m}{8834s} = 625900 m{s}^{-1}[/math] [math]\frac{2\times\pi\times8.8\times10^8 m}{8834s} = 625900 m{s}^{-1}[/math] Edited July 12, 2011 by murshid 0 Share this post Link to post Share on other sites

csmyth3025 27 Posted July 12, 2011 [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] [ 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 0 Share this post Link to post Share on other sites

Cap'n Refsmmat 1351 Posted July 12, 2011 You can also use \mbox{} for the m/s. Click on the equation to see the code used to make it: [math]625,900 \, \mbox{m/s}[/math] 0 Share this post Link to post Share on other sites

Quicksdk 0 Posted March 22, 2012 Please, help me install IPBlatex on IPBoard. If I want to install IPBLatex, I should have any program on my computer? 0 Share this post Link to post Share on other sites

amirhomayoun 0 Posted November 25, 2012 just testing $A \bigcup B$ 0 Share this post Link to post Share on other sites

StringJunky 2289 Posted November 25, 2012 (edited) just testing $A \bigcup B$ You can test here: http://www.scienceforums.net/topic/58931-latex-testing/ Edited November 25, 2012 by StringJunky 0 Share this post Link to post Share on other sites

DevilSolution 26 Posted March 25, 2013 (edited) [math] e = mc^2 [/math] Edited March 25, 2013 by DevilSolution 0 Share this post Link to post Share on other sites

kmath 0 Posted April 7, 2013 How do I write matrices in latex, ie a b c d ? 0 Share this post Link to post Share on other sites

imatfaal 2481 Posted April 8, 2013 \left( \begin{array}{ccc}a & b & c \\d & e & f \\g & h & i \end{array} \right) [latex]\left( \begin{array}{ccc}a & b & c \\d & e & f \\g & h & i \end{array} \right)[/latex] There is probably an easier way 0 Share this post Link to post Share on other sites

Nehushtan 3 Posted April 8, 2013 You can also try: \begin{pmatrix}a & b & c \\d & e & f \\g & h & i \end{pmatrix} [latex]\begin{pmatrix}a & b & c \\d & e & f \\g & h & i \end{pmatrix}[/latex] 1 Share this post Link to post Share on other sites

nilesh dahiya 0 Posted April 17, 2014 like this? /neq 0 Share this post Link to post Share on other sites

imatfaal 2481 Posted April 17, 2014 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 [latex]\neq[/latex] 0 Share this post Link to post Share on other sites

knyazik 7 Posted October 22, 2014 Thank you so much. This guide was incredibly useful to me! 0 Share this post Link to post Share on other sites

MonDie 145 Posted November 14, 2014 You didn't include overbar! \bar{x} [math]\bar{x}[/math] But this was the first thing I couldn't find in the tutorial. nice job 0 Share this post Link to post Share on other sites

knyazik 7 Posted November 17, 2014 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? 0 Share this post Link to post Share on other sites

DevilSolution 26 Posted November 16, 2015 (edited) [math]b_n= \frac{1}{T} \int^{T}_{0} f(t) \sin( \frac{2 \pi nt}{T})dt[/math] [math]x^2[/math] [math]xt=a0+\sum[a_k * \cos(\lambda_k * t) + b_k*\sin(\lambda_k*t)] [/math] [math]\lambda = 2 * \pi * \nu_k[/math] [math]\nu_k = k/q[/math] [math]a_0 = \frac{1}{T} \int^{T}_{0} f(t) dt [/math] [math]\frac{cov}{\sum^{n}_{i = 1}(x_i - \bar{x})^2 } [/math] [math]\sum^{n}_{i = 1}[/math] Edited November 16, 2015 by DevilSolution 0 Share this post Link to post Share on other sites

DevilSolution 26 Posted November 19, 2015 (edited) [math]70*\tn[/math] [math] [/math] Edited November 19, 2015 by DevilSolution 0 Share this post Link to post Share on other sites

LaurieAG 13 Posted November 19, 2015 [math] {\bar \lambda} = \frac {\lambda}{2 \pi} = \frac{\hbar}{m c}[/math] 0 Share this post Link to post Share on other sites

renerpho 9 Posted September 28, 2016 (edited) It works Edited September 28, 2016 by renerpho 0 Share this post Link to post Share on other sites

renerpho 9 Posted September 29, 2016 (edited) Notice that your sum is of the form [math]\sum_{n=1}^{m}\frac{P(n)}{2^n}[/math] where [math]P[/math] is a polynomial. You start your proof with an educated guess, that the sum will be of similar form, namely [math]\sum_{n=1}^{m}\frac{n^3}{2^n}\stackrel{?}{=}\frac{{Q{_1}(m)}2^m+Q{_2}(m)}{2^m}[/math] where [math]Q{_1}[/math] and [math]Q{_2}[/math] are themselves polynomials (you include a polynomial term multiplied by [math]2^m[/math] to increase your chances of success). You can be quite confident that [math]Q{_2}[/math] will be of at least same degree as [math]P[/math]. So, your first attempt is the simplest possible, where [math]Q{_1}[/math] has degree 0 (turning it into a constant) and [math]Q{_2}[/math] has degree 3. This leads you to [math]\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}[/math] for some real numbers [math]a,b,c,d,e[/math]. Evaluate at [math]m=1,\cdots,5[/math] 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 September 29, 2016 by renerpho 0 Share this post Link to post Share on other sites

Sriman Dutta 49 Posted October 23, 2016 testing [math] s^{/frac{1}{2}} = 0 [/math] 0 Share this post Link to post Share on other sites

imatfaal 2481 Posted October 23, 2016 testing [math] s^{/frac{1}{2}} = 0 [/math] your slash before the frac is the wrong way around [latex] s^{\frac{1}{2}} [/latex] click on mine to see difference 0 Share this post Link to post Share on other sites

Sriman Dutta 49 Posted October 23, 2016 Oh I see. thanks. 0 Share this post Link to post Share on other sites