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# Typing math

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I've added support for typesetting math using LaTeX on the forums.

Example:

$$\sum_{n = 0}^\infty \frac{1}{n!} = \lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n = e$$

Outputs:

$$\sum_{n = 0}^\infty \frac{1}{n!} = \lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n = e$$

Just in case anyone wanted to write some math that's a little more advanced than basic arithmetic.

Update:

fear my skills.

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$$1+1=2$$

fear my skills.

$$1+1=3$$

That's how I roll.

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$$\widetilde{\underline{\zeta\Gamma\alpha\gamma\sigma}}\sqrt[n]{\underline{|\phi}\int\underline{\Sigma\eta\tau}}$$

That's how I roll.

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Yeah, that's fine but...

$$-1 = 1$$

$$-1 = i \cdot i = \sqrt{-1} \cdot \sqrt{-1} = \sqrt{-1 \cdot -1} = \sqrt{1} = 1$$

I win...

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Except

$$i \not = \sqrt{-1}$$

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My calculator says otherwise...

See attachment... Stupid Texas Instruments didn't bother creating an x64 driver

[attachment deleted by admin]

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Well... the definition is $$x^2 = -1$$

Saying that $$i = \sqrt{-1}$$ is not incorrect as long as you make it clear, that $$\sqrt$$ operator is redefined here as compared to the one used for real numbers.

$$\sqrt{-x} = i \sqrt x$$ where $$0 < x \in \mathbb{R}$$

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$$x^n = y \Leftrightarrow \sqrt[n]{x^n} = x = \sqrt[n]{y}$$, no?

It's because $$\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$$ is only true for $$a \in \Bbb{R}^+ \cup \{0\}$$ and $$b \in \Bbb{R}^+ \cup \{0\}$$ 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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Well, looks as though two people had reasons to use the latex addon.  lol

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$$0+0=1$$

Thats how I roll!

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$$0+0=1$$

Thats how I roll!

My mums Excel spreadsheets do this kind of thing all the time.

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$$0.\overline{999} = 1$$

$$\frac3{9} = 0.\overline{333}$$

$$\frac6{9} = 0.\overline{666}$$

$$so$$

$$\frac9{9} = 0.\overline{999}$$

$$but$$

$$\frac9{9} = \frac1{1} = 1$$

Lol.

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And that's why patterns shouldn't be assumed based off of 2 terms ;p.

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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:

$$\frac{1}{3} = 0.\overline{333}$$

$$3 \cdot 0.\overline{333} = 3 \cdot \frac{1}{3} = \frac{3}{3} = 1$$

Therefore:

$$0.\overline{999} = 1$$

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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....

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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:

$$.9 = \frac{9}{10^1}$$

$$.09 = \frac{9}{10^2}$$

$$.009 = \frac{9}{10^3}$$

etc.

$$.9 + .09 + .009 = \frac{9}{10^1} + \frac{9}{10^2} + \frac{9}{10^3} = .999$$

So:

$$\sum_{n=1}^{\infty} \frac{9}{10^n} = 0.\overline{999} = 1$$

You see that as $$n \to \infty$$ (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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Wouldn't .999 (repeating) = 1 - 1/inf, not 1?

Yeah, but that's the same thing because $$\frac{x}{\infty} = 0$$ and 1 - 0 = 1

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Wait....  x/inf is always assumed to be 0?

Ok then...  I shall now shut up ;p.

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