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:

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

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

Therefore:

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


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…

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Jan 3 2012

Snow White

Surely not this ugly girl...

For those of you who doesn’t know Linear Algebra (yet…), the matrix on the blackboard is a reflection matrix, when multiplying with a \mathbb R^{2} vector, the matrix will change the sign of the x coordinate, therefor it will create a reflected image of the vector.