Wednesday, February 29, 2012

Abbreviated Journal Names in Mendeley | alex m. chubaty

Abbreviated Journal Names in Mendeley | alex m. chubaty

This might be useful to some. This blog post is about Mendeley and using abbreviated journal names. Some journal publications use the abbreviated journal names in their bibliographies (references). One such journal is the Journal of Fluid Mechanics.

How to Install Elementary Luna OS in Ubuntu 12.04 Using this Simple Script | Tech Drive-in

How to Install Elementary Luna OS in Ubuntu 12.04 Using this Simple Script | Tech Drive-in

Alright, I'm back! Passed the defense, woohoo!

This is an interesting article on Elementary "Luna" OS built upon Ubuntu 12.04. Still alpha version so is very unpredictable.

A quote from the article:
Other major apps that comes as default with Elementary OS Luna include BeatBox Music Player, Maya calendar and organizing tool, Marlin file browser, Footnote note taking application, Feedler feed reader, Midori web browser, Plank dock application, Snap for taking pictures, Postler email client, Dexter address book and so on. As you can see, folks behind Elementary Project are trying to create a lean mean Linux distro with a clear focus on detail and consistency. I hope things pan out well for them. Good luck to the devs.

Thursday, February 23, 2012

10 New Features Added to Ubuntu 12.04 Precise Pangolin | Tech Drive-in

10 New Features Added to Ubuntu 12.04 Precise Pangolin | Tech Drive-in

Sorry about the lack of activity, I've been busy with a defense presentation. Here is an interesting read on some updates about the next Ubuntu, 12.04

Tuesday, February 14, 2012

Friday, February 10, 2012

Sage ImportError: No module named _md5 - Ubuntu 11.10

Ok, so I installed a fresh copy of Ubuntu 11.10 on a few Windows machines at school and downloaded Sage 4.8 to use.  However, I was getting the error:

ImportError: No module named _md5

when trying to run Sage.  The fix for this is to install the package:


by either executing in the terminal the command:

sudo apt-get install libssl0.9.8

or installing it from the USC.

Wednesday, February 8, 2012

Ubuntu 11.10 - update on a few things from me

Ok, so I installed Ubuntu 11.10 on a few of my computers at school and have noticed a few things.

First, it seems that Ubuntu finally updated its repositories in the USC to reflect the latest wxMaxima version which is 11.04.

I also noticed that is seems synaptic package manager is slowly becoming obsolete.  I could install it, but I couldn't run it while the USC was open.  So I decided to install my wanted extra TeX Live packages from the USC instead.  It worked pretty well.

I am also cutting down on software packages, especially for math/science.  I have cut off the simple graphing and calculating software such as KAlgebra, Libniz, Euler, etc., and I am sticking mainly with wxMaxima and Sage.  I am also trying to get into Python.

Marlin File Browser is Evolving Pretty Nicely | Tech Drive-in

Marlin File Browser is Evolving Pretty Nicely | Tech Drive-in

For Ubuntu. Not official yet, though.

Friday, February 3, 2012

Natural log and trig relation: -ln(tan(phi/2)) = ln(csc(phi) + cot(phi)) and the integrals of csc^3(phi) and csc(phi)


- \ln \left( \tan \dfrac{\phi}{2} \right) = - \ln \left( \dfrac{\sin \phi}{1 + \cos \phi} \right) = \ln \left( \dfrac{1 + \cos \phi}{\sin \phi} \right) = \ln \left( \dfrac{1}{\sin \phi} + \dfrac{\cos \phi}{\sin \phi} \right)


= \ln \left( \csc \phi + \cot \phi \right)


However, I recently was using this in an analysis, was checking some work, and noticed that:


\ln \left( \tan \dfrac{\phi}{2} \right) = \ln \left( \dfrac{1 - \cos \phi}{\sin \phi} \right) =  \ln \left( \csc \phi - \cot \phi \right)


I looked up the formulas again and became quite confused.

The tangent half angle relations are:

\[ \tan \dfrac{\phi}{2} = \dfrac{1 - \cos \phi}{\sin \phi}\]


\[ \tan \dfrac{\phi}{2} = \dfrac{\sin \phi}{1 + \cos \phi}\]

So, I guess the relation is:

\[ \ln \left( \csc \phi - \cot \phi \right) = -\ln \left( \csc \phi + \cot \phi \right) \]

Yup, it is confirmed:

Links to the tangent half angle:

However, my problem originates to the integral of \( \csc^3 \phi \).  This integral can be solved by (at least how I did it)

\[ \int \csc^3 \phi \mathrm{d} \phi \]

First, utilize integration by parts

\[ \mathrm{d}v = \csc^2 \phi \mathrm{d} \phi \]

\[ u = \csc \phi \]

\[ v = -\cot \phi \]

\[ \mathrm{d}u = - \csc \phi \cot \phi \mathrm{d} \phi \]

The formula for integration by parts is

\[ uv - \int v \mathrm{d}u \]

Substituting in the correct values gives

\[ -\csc \phi \cot \phi - \int \csc \phi \cot^2 \phi \mathrm{d} \phi \]

Using the trigonometric identity \( \cot^2 \phi = \csc^2 \phi - 1 \) transforms the relation to

\[ -\csc \phi \cot \phi - \int \csc \phi \left( \csc^2 \phi - 1 \right) \mathrm{d} \phi \]

Multiplying through we obtain

\[ -\csc \phi \cot \phi - \int \csc^3 \phi - \csc \phi \mathrm{d} \phi \]

Separating the RHS integrals produces

\[ -\csc \phi \cot \phi - \int \csc^3 \phi \mathrm{d} \phi +  \int \csc \phi \mathrm{d} \phi \]

Recall that this whole relation is equal to \(  \int \csc^3 \phi \mathrm{d} \phi \) which yields

\[  \int \csc^3 \phi \mathrm{d} \phi = -\csc \phi \cot \phi - \int \csc^3 \phi \mathrm{d} \phi +  \int \csc \phi \mathrm{d} \phi \]

At first this seems counter-intuitive and does not seem to help.  However, if you collect the integrals of \( \csc^3 \phi \) the equation becomes clearer as where to go next.  So collecting the \( \csc^3 \phi \) integrals to the LHS, we obtain

\[  2 \int \csc^3 \phi \mathrm{d} \phi = -\csc \phi \cot \phi +  \int \csc \phi \mathrm{d} \phi \]

Then we can divide by two and evaluate the left hand side which gives us

\[  \int \csc^3 \phi \mathrm{d} \phi = \dfrac{1}{2} \left[ -\csc \phi \cot \phi +  \int \csc \phi \mathrm{d} \phi \right] \]

Okay, so this is where my problem began.  I've found two answers for this integral of \( \csc \phi \).  The first is the same as the one I got by evaluating the integral of \( \csc \phi \)

\[  \int \csc^3 \phi \mathrm{d} \phi = \dfrac{1}{2} \left[ -\csc \phi \cot \phi - \ln | \csc \phi - \cot \phi | \right] \]


\[ \int \csc \phi \mathrm{d} \phi = - \ln | \csc \phi - \cot \phi | \]

according to:

However, here:, the answer is

\[ \int \csc \phi \mathrm{d} \phi = - \ln | \csc \phi + \cot \phi | \]

?? (I'll have to check these to make sure there isn't a mistake.

To make matters worse here: we get an answer for \( \int \csc^3 \phi \mathrm{d} \phi \) equal to

\[  \dfrac{1}{2} \left[ -\csc \phi \cot \phi + \ln | \csc \phi - \cot \phi | \right] \]

??? What the heck is going on?! (Again, when I get the time I will check these answers since the formulation is given.  However, this does give the final answer I am looking for which is

\[  \dfrac{1}{2} \left \lbrace -\csc \phi \cot \phi + \ln \left[ \tan \left( \dfrac{\phi}{2} \right) \right] \right \rbrace \]

However again, the relation I first displayed above shows that

\[  \dfrac{1}{2} \left[ -\csc \phi \cot \phi - \ln | \csc \phi + \cot \phi | \right] \]

\[ = -\dfrac{1}{2} \left[ \csc \phi \cot \phi + \ln | \csc \phi + \cot \phi | \right] \]

\[ = -\dfrac{1}{2} \left[ \csc \phi \cot \phi - \ln | \csc \phi - \cot \phi | \right] \]

\[ = \dfrac{1}{2} \left[ -\csc \phi \cot \phi + \ln | \csc \phi - \cot \phi | \right] \]

So this tells me that the second link should be correct!  The reason for this post originates to me trying to change the angle \( \phi \) from spherical polar coordinates to cylindrical polar coordinates \( \left( r, z \right) \) through the relation \( \phi = \tan \left( r/z \right) \)and \( R^2 = r^2 + z^2 \).  Thus, I am posting this to keep myself from become confused!