Drawing Blanks

Premature Optimization is a Prerequisite for Success

Obvious identity

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Marquis de Laplace said that those to whom this identity is not obvious are clueless in math:

\int_{-\infty}^{\infty} e^{-x^2}dx=\sqrt{\pi}

Seriously, come on:

\int\int e^{-y^2}e^{-x^2}dx dy = \int_0^{\infty} e^{-r^2} 2\pi r dr =\pi\int_0^{\infty}e^{-r^2}d(r^2)=\pi

(just testing latex notation)

Another try

\int \limits_{-\infty}^{+\infty}\int \limits_{-\infty}^{+\infty}e^{-(x^2+y^2)}dxdy=\int \limits_0^{2{\pi}}\int \limits_0^{+\infty} e^{-r^2}rdr d \varphi =\pi \int \limits_0^{+\infty} e^{-r^2}d(r^2)=\pi


Written by bbzippo

09/19/2011 at 6:58 am

Posted in math

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