Tag Archives: numbers

The beauty of Mayavi

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Four charges with different magnitude plotted in 3D using Mayavi

In one of my earlier posts about Mayavi, I wrote about how you could visualize 2D field line plots using the flow function. At the end of that post I added that Mayavi is actually best at 3D plotting, and to follow up on that I’ll show you some of these plots with a few example Python scripts you might try out on your own.

First of all, you might want to know how to install Mayavi. For those lucky ones of you who have freed yourself and jumped on the Linux bandwagon, installing Mayavi should be quite easy. If you are using Ubuntu in particular, you may just install the package mayavi2 using either Synaptic or apt-get. If you are on Windows or Mac, you may either install Enthought’s own Python distribution (EPD) or give a shot at compiling on your own. Just note that EPD is quite expensive, even though all its components are open source, but if you are a student or academic user you could go ahead and download the academic version for free. It is basically the same as the commercial one, but with an academic license. (Kudos to Enthought for both making Mayavi open source, building an business model around it and still providing a great solution for students!)

Now, Enter 3D!

The way you do your plots in Mayavi depends on what you want to express. Most likely, you would prefer to show some simple plots giving just the necessary amount of information to tell you how the electric field behaves around your charges. A simple example of this is shown below:

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Sweet first proof

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Carl Friedrich Gauss (1777 – 1855) by G. Biermann (1824-1908)

Earlier today, enjoying a warm cup of coffee with a friend of mine,we got into a discussion about some math and ended up contemplating on how to prove the sum formula for the n first natural numbers.

The proof is rumored to first have been done by Gauss when he was only a child.

Let S_n = 1 + 2 + 3 + . . . + (n-2) + (n-1) + n

just rewriting the terms backwards we get

S_n = n + (n-1) + (n-2) + \ldots + 3 + 2 +1

now adding these two expressions for the sum we obtain

2 S_n = (n+1) + (n+1) + (n+1) + \ldots+ (n+1) + (n+1) + (n+1)

and since we had n terms in the original sum we now have n \cdot (n+1)‘s, so

2 S_n = (n+1) + (n+1) + (n+1) + \ldots+ (n+1) + (n+1) + (n+1) = n(n+1)

so..

S_n = \frac{ n(n+1) }{2}

Neat. On a computer this would severely reduce the number of operations that would have to be done to compute such a sum. Imagine having to sum up the first million natural numbers and let’s suppose the computer requires one operation for adding, multiplying, dividing and so forth. Then implementing this formula would reduce the numbers of operations from 10^6 to 3.