Copasetic Flow

You may have noticed that I left you hanging with the arctangent series a few weeks ago.  I told you how to get the series for arctan for x > 1, but not for x < 1.  By now, I bet my EM grader has noticed as well, and I've lost a few points.  So, a little after the fact, here's how to get a series for arctangent for x < 1. The useful bit you need that I didn't have is that the arctan of x can also be expressed as (picture 1): The following picture, (picture 2), has the geometric interpretation that helped me see things more easily as well as the required series.  If you think of the tangent of an angle being equal to the opposite side divided by adjacent side, and then take a look at the lower left corner of the picture below, I think you'll see why the expression above is true.  It took forever for me to see it after my  professor derived it analytically.  I think I could have seen it much more quickly derived geometrica...

I'm getting ready to present my superconductor research [1] at the Texas Academy of Science meeting this weekend, so today's post is just a quick note on an easy way to find a series expansion for the arctangent. Arctan(x) looked a little daunting to get into a series at first.  The key to get started is to remember what the derivative of arctan is (picture 1) At least the derivative is something that looks like it could be pretty easily worked into a series.  With a little more thought, it turns out that the anti-derivative, (the integral), of the result above is arctan.  So, the process will be to turn the derivative into a series first, and then take the integral of that series to wind up with the series for arctan. The result of the derivative shown above is a prime candidate for a geometric series expansion(picture 2) Now that we have the series, all that remains is to integrate and we get (picture 3): So (picture 4), References: 1.   ht...

Yeah for Google+!!!  After posting yesterday's missive about figuring out the origin of the secant squared term, mathematician extraordinare and Google+er  +John Baez  pointed out to me that the derivative of tangent is secant squared and that I need not have fussed so much with the geometry.  John said: I wouldn't  call the introduction of sec^2(θ) a "substitution".  Introducing θ in the first place counts as a substitution, since you're trying to simplify an integral by replacing some other variable with this one.   But it's just a mathematical fact that the derivative of tanθ is sec^2(θ), so d tan(θ) = sec^2(θ) dθ I would use this equation automatically and unthinkingly, but it looks like you're trying to find a geometrical explanation of this equation.  If you're figuring it out for yourself because nobody told you the derivative of tan(θ) is sec^2(θ), that's very laudable!  The standard approach is to express tan in terms of sin and...

In studying form my EM midterm, I came across a practice problem involving finding the potential along the z axis due to a charged disc centered on the same axis.  After thinking about the problem a bit, I turned to what's becoming one of my favorite online references, Dr. J.B. Tatum's text on electricity and magnetism [2].  Sure enough, there was a solution that could be adapted to the task at hand.  It involved several 'clever' trig substitutions one of which were not immediately clear to me, so I've expanded upon it here.  After the clever trig trick, read on to find out more interesting stuff about Dr. Tatum, an emeritus professor at the University of Victoria [1]. The Basic Problem The practice problem mentioned above is described by the following diagram from Dr. Tatum's text.  The first and handiest innovation in Dr. Tatum's treatment is to parameterize the problem using the angle marked as theta and the limit of that angle labeled as alpha. T...