Jan 18 2012

Easy way to bypass Wikipedia’s Blackout

As most of you know, Wikipedia is protesting against SOPA and PIPA today, so they blocked the English Wikipedia. Which can be kind of annoying.

I’ve found a quick workaround that allows you to use Wikipedia like every normal day. And I’ve build a small software to apply the workaround. All it does is patching the host file, and changing one server (It has an unpatch button as well, in order to remove the patch and getting back to the old settings).

Just download it and click patch!

Mirror #1

Jan 18 2012

Gaussian Integral

A Gaussian function (named after one of the greatest mathematician, Carl Friedrich Gauss) is a function of the form:

    \[ f\left(x\right)=ae^{-\frac{\left(x-b\right)^{2}}{2c^{2}}} \]

We would have like to examine the following integral:

    \[ I=\intop_{-\infty}^{\infty}e^{-x^{2}}\mathrm{d} x\]

But we are facing a problem, what is the antiderivative function of e^{-x^{2}}?

Well, You won’t be able to find one, at least not an elementary one. The integral of a Gaussian is the Gauss error function, but in this special case, there is a cool way to calculate this integral.

At first, Instead of looking at I, lets take a look at I^{2}. Meaning:

    \[ I^{2}=\intop_{-\infty}^{\infty}e^{-x^{2}}\mathrm{d} x\intop_{-\infty}^{\infty}e^{-y^{2}}\mathrm{d} y \]

Note: In the second integral, I’ve changed the variable of integration to y. I can do that, because it’s a completely separated integral.

Now, notice that the following integral is equivalent to the following:

    \[ \intop_{-\infty}^{\infty}\intop_{-\infty}^{\infty}e^{-x^{2}-y^{2}}\mathrm{d} x\mathrm{d} y \]

Well, That’s a bit weird. Now we have a double integral! We’ve made the problem much more complex! Or did we?

Well, yes, now we have a double integral, but there is a good reason for that. Now we can change our coordinate system to the polar coordinate system (Meaning: x=r\cos\theta, y=r\sin\theta) and remember to multiply by the Jacobian determinant! In our case, the Jacobian determinant is r. So, we got:

    \[ \intop_{0}^{2\pi}\intop_{0}^{\infty}e^{-r^{2}}r\mathrm{d}r\mathrm{d}\theta \]

But wait a minute, this looks familiar! Notice that:

    \[ =-\frac{1}{2}\intop_{0}^{2\pi}\intop_{0}^{\infty}\frac{\mathrm{d}}{\mathrm{d}r}\left(e^{-r^{2}}\right)\mathrm{d}r\mathrm{d}\theta=-\frac{1}{2}2\pi\intop_{0}^{\infty}\frac{\mathrm{d}}{\mathrm{d}r}\left(e^{-r^{2}}\right)\mathrm{d}r=\]

    \[ =- \pi \left[e^{-r^{2}}\right]_{0}^{\infty}= -\pi\left[0-1\right]=\pi\]


    \[ I^{2}=\pi\Rightarrow I=\intop_{-\infty}^{\infty}e^{-x^{2}}\mathrm{d} x = \sqrt{\pi} \]

Jan 9 2012

Lord of the Rings

One Ring to fail them all...

Today we’ve started learning about Rings. That was the first thing that popped into my head. Yeah, I know, a bit lame…