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Thread: Different ways of calculating pi

  1. #31
    MHF Contributor Quick's Avatar
    May 2006
    New England
    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; Dec 19th 2006 at 01:15 PM.
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