Comments by Paul

  • Math on the Pololu website!

    - 27 June 2014

    Ben wanted me to post the details of the Gaussian integral, so here we go.

    It's simpler to think about the full integral from ``-oo`` to ``+oo``. Let's give that a name:

    ``lambda = int_-oo^oo e^(-x^2) dx``.

    Then, instead of trying to compute ``lambda``, we work on ``lambda^2``:

    ``lambda^2 = ( int_-oo^oo e^(-x^2) dx )^2 = int_-oo^oo int_-oo^oo e^(-x^2) e^(-y^2) dx dy``.

    We note that ``e^(-x^2) e^(-y^2) = e^(-(x^2+y^2))`` and rewrite in polar coordinates:

    ``lambda^2 = int_0^oo e^(-r^2) 2pi r \ dr``.

    Luckily, that integral is easy, since ``d / (dr) pi e^(-r^2) = - 2 pi r \ e^(-r^2)``:

    ``lambda^2 = - [ pi e^(-r^2) ]_0^oo = pi``;

    ``lambda = sqrt(pi)``.

    Finally, since the Gaussian is an even function, our desired integral is just half of that value:

    ``int_0^oo e^(-x^2) dx = sqrt pi / 2``.

  • Pocket-sized USB charger adapter

    Pocket-sized USB charger adapter

    - 6 March 2014

    Thanks for the nice feedback, and please let us know when you post more projects!

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