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Typing math


Daniel0

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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.

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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]

 

 

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[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.

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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]

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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.

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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.)

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