yo

[Ender]

you need to change your forum description from this "3.141592654"

to this
[tex]2\int_{-1}^{1} \sqrt{1-x^2}dx[/tex]

or at least to this "3.141592654..."

GET WTIHT HE PROGRAM.

[Joe]

4 * (1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11...)

What's the proper way of writing that continuing sequence?

[Ender]

4 * (1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + ...) is more conventional

you get this from pi/4 = arctan(1), and then the sum 1 - 1/3 + 1/5 + ... is the taylor series for arctan evaluated at 1

but pi is often defined as the integral that i posted, whereas this other representation is derived

speaking of which, it's cool how cos is defined:

[tex]for \; -1 \leq x \leq 1 \; let \;A(x) = x\sqrt{1-x^2} + 2\int_{x}^{1} \sqrt{1-t^2}dt[/tex]

then we define

[tex]\cos = A^{-1}[/tex]

but you first have to show that A is a bijection (it's sufficient to show that it's strictly decreasing)

and then sin is defined as

[tex]\sin x = \sqrt{1 - \cos^2 x}[/tex]