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Math Help - Different ways of calculating pi

  1. #31
    MHF Contributor Quick's Avatar
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    Just to remind you, my method uses extremely simple geometry to create pi, in fact, it only uses the pythagorean theorum
    Quote Originally Posted by TriKri View Post
    I dont get this ... as far as I can see, you calculate y' from the equation y = \sqrt{1-x^2}, to put an integral on y', to get ... pi? How is that...

    Isn't it simpler just integrating y directly:

    \sum_{a=0}^n \frac{\sqrt{1-\left(\displaystyle\frac{a}{n}\right)^2}}{n}\ =\ \frac{1}{n}\sum_{a=0}^n \sqrt{\frac{n^2-a^2}{n^2}}\ = \ \frac{1}{n^2}\sum_{a=0}^n \sqrt{n^2-a^2}
    Just for your information, I'm currently in "freshman geometry" in high-school, I haven't reached calculus
    (I must have fooled you )


    And one thing I spoted is that you set \Delta y = \sqrt{\Delta y^2}, that's equal to |\Delta y| for real numbers \Delta y.
    The way I set up the problem, \Delta y is always negative. However, since the pythagorean theorum uses \Delta y^2 I decided it didn't matter
    Last edited by Quick; December 19th 2006 at 02:15 PM.
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