Daniel0 Posted January 9, 2009 Share Posted January 9, 2009 I've added support for typesetting math using LaTeX on the forums. Example: [tex]\sum_{n = 0}^\infty \frac{1}{n!} = \lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n = e[/tex] Outputs: [tex]\sum_{n = 0}^\infty \frac{1}{n!} = \lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n = e[/tex] Just in case anyone wanted to write some math that's a little more advanced than basic arithmetic. Update: The [tex] bbcode is now deprecated in favor of [math] and [imath]. The former is used for display mode, the latter is for inline mode. Example of display mode: [math]\sum_{n = 0}^\infty \frac{1}{n!} = \lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n = e[/math] Inline mode is used for math in sentences, like you might want to say [imath]\frac{1+2}{5}\cdot7=\frac{21}{5}[/imath] within a sentence. Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/ Share on other sites More sharing options...
Mchl Posted January 11, 2009 Share Posted January 11, 2009 Is this a released SMF mod, or something you created? I could use it on my forum. Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-734454 Share on other sites More sharing options...
Daniel0 Posted January 11, 2009 Author Share Posted January 11, 2009 It's this one: http://custom.simplemachines.org/mods/index.php?mod=1111 Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-734607 Share on other sites More sharing options...
Mchl Posted January 11, 2009 Share Posted January 11, 2009 Thanks Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-734613 Share on other sites More sharing options...
.josh Posted January 11, 2009 Share Posted January 11, 2009 [tex]1+1=2[/tex] fear my skills. Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-734918 Share on other sites More sharing options...
Mchl Posted January 11, 2009 Share Posted January 11, 2009 Splendid! Here's your skill badge! Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-734920 Share on other sites More sharing options...
corbin Posted January 11, 2009 Share Posted January 11, 2009 [tex]1+1=2[/tex] fear my skills. [tex]1+1=3[/tex] That's how I roll. Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-734925 Share on other sites More sharing options...
.josh Posted January 11, 2009 Share Posted January 11, 2009 [tex]\widetilde{\underline{\zeta\Gamma\alpha\gamma\sigma}}\sqrt[n]{\underline{|\phi}\int\underline{\Sigma\eta\tau}}[/tex] That's how I roll. Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-734934 Share on other sites More sharing options...
Daniel0 Posted January 11, 2009 Author Share Posted January 11, 2009 Yeah, that's fine but... [tex]-1 = 1[/tex] [tex]-1 = i \cdot i = \sqrt{-1} \cdot \sqrt{-1} = \sqrt{-1 \cdot -1} = \sqrt{1} = 1[/tex] I win... Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-734937 Share on other sites More sharing options...
Mchl Posted January 11, 2009 Share Posted January 11, 2009 Except [tex] i \not = \sqrt{-1}[/tex] Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-734942 Share on other sites More sharing options...
Daniel0 Posted January 11, 2009 Author Share Posted January 11, 2009 My calculator says otherwise... See attachment... Stupid Texas Instruments didn't bother creating an x64 driver [attachment deleted by admin] Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-734947 Share on other sites More sharing options...
Mchl Posted January 11, 2009 Share Posted January 11, 2009 Well... the definition is [tex]x^2 = -1[/tex] Saying that [tex] i = \sqrt{-1}[/tex] is not incorrect as long as you make it clear, that [tex] \sqrt[/tex] operator is redefined here as compared to the one used for real numbers. [tex]\sqrt{-x} = i \sqrt x[/tex] where [tex]0 < x \in \mathbb{R} [/tex] Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-734952 Share on other sites More sharing options...
Daniel0 Posted January 11, 2009 Author Share Posted January 11, 2009 [tex]x^n = y \Leftrightarrow \sqrt[n]{x^n} = x = \sqrt[n]{y}[/tex], no? It's because [tex]\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}[/tex] is only true for [tex]a \in \Bbb{R}^+ \cup \{0\}[/tex] and [tex]b \in \Bbb{R}^+ \cup \{0\}[/tex] or something like that. That's the "trick" or whatever... There is another "proof" that says that any integer a equals any other integer b, I just can't remember that one, but there is of course an error in that one as well. Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-734958 Share on other sites More sharing options...
corbin Posted January 12, 2009 Share Posted January 12, 2009 Well, looks as though two people had reasons to use the latex addon. lol Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-735029 Share on other sites More sharing options...
Mchl Posted January 12, 2009 Share Posted January 12, 2009 http://en.wikipedia.org/wiki/False_proof Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-735116 Share on other sites More sharing options...
aznkidzx Posted March 1, 2009 Share Posted March 1, 2009 [tex]0+0=1[/tex] Thats how I roll! Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-773565 Share on other sites More sharing options...
Mchl Posted March 1, 2009 Share Posted March 1, 2009 [tex]0+0=1[/tex] Thats how I roll! My mums Excel spreadsheets do this kind of thing all the time. Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-773726 Share on other sites More sharing options...
ShadeSlayer Posted May 10, 2009 Share Posted May 10, 2009 [tex]0.\overline{999} = 1[/tex] [tex]\frac3{9} = 0.\overline{333} [/tex] [tex]\frac6{9} = 0.\overline{666} [/tex] [tex]so[/tex] [tex]\frac9{9} = 0.\overline{999}[/tex] [tex]but[/tex] [tex]\frac9{9} = \frac1{1} = 1[/tex] Lol. Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-830728 Share on other sites More sharing options...
corbin Posted May 10, 2009 Share Posted May 10, 2009 And that's why patterns shouldn't be assumed based off of 2 terms ;p. Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-830808 Share on other sites More sharing options...
Daniel0 Posted May 10, 2009 Author Share Posted May 10, 2009 Actually corbin, it's widely accepted among mathematicians that .999... = 1. ShadeSlayer's proof may be said to be flawed in the sense that it doesn't explain how the pattern he uses works. A better version might be: [tex]\frac{1}{3} = 0.\overline{333}[/tex] [tex]3 \cdot 0.\overline{333} = 3 \cdot \frac{1}{3} = \frac{3}{3} = 1[/tex] Therefore: [tex]0.\overline{999} = 1[/tex] Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-830823 Share on other sites More sharing options...
corbin Posted May 10, 2009 Share Posted May 10, 2009 Hrmmm.... Yeah.... Strange to me though since .999 (repeating) would never actually equal 1. Then again, if I look at it that way, repeating .333 would never actually equal 1/3.... Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-831078 Share on other sites More sharing options...
Daniel0 Posted May 10, 2009 Author Share Posted May 10, 2009 Dunno if it helps your understanding, but 0.99 is closer to 1 than 0.9 is, so for each 9 you add, you get closer to 1. If you have an infinite amount of 9's then you are getting infinitely close to 1. Or if you want it in math terms: [tex].9 = \frac{9}{10^1}[/tex] [tex].09 = \frac{9}{10^2}[/tex] [tex].009 = \frac{9}{10^3}[/tex] etc. [tex].9 + .09 + .009 = \frac{9}{10^1} + \frac{9}{10^2} + \frac{9}{10^3} = .999[/tex] So: [tex]\sum_{n=1}^{\infty} \frac{9}{10^n} = 0.\overline{999} = 1[/tex] You see that as [tex]n \to \infty[/tex] (analogous to "you are adding more 9's on the end") you are getting closer to 1. Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-831082 Share on other sites More sharing options...
corbin Posted May 10, 2009 Share Posted May 10, 2009 Yeah, we actually discussed this in class one day when talking about geometric sequences a while back. And yes, as n -> inf., the number gets closer to 1, but it would never actually reach one. Wouldn't .999 (repeating) = 1 - 1/inf, not 1? (I do realize that obviously I'm arguing pointlessly since people accept .999 to be 1, just like .333 is assumed to be 1/3 [both repeating]. I guess I just think that things should be left in fractional form ;p.) Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-831085 Share on other sites More sharing options...
Daniel0 Posted May 10, 2009 Author Share Posted May 10, 2009 Wouldn't .999 (repeating) = 1 - 1/inf, not 1? Yeah, but that's the same thing because [tex]\frac{x}{\infty} = 0[/tex] and 1 - 0 = 1 Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-831086 Share on other sites More sharing options...
corbin Posted May 10, 2009 Share Posted May 10, 2009 Wait.... x/inf is always assumed to be 0? Ok then... I shall now shut up ;p. Quote Link to comment https://forums.phpfreaks.com/topic/140190-typing-math/#findComment-831090 Share on other sites More sharing options...
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