Complex Analysis

Posted by tpc at April 30th, 2008

I’ve finally wiped the dust off my copy of Ahlfors and started reading from page 1. Previously, I’ve only looked at the chapter on Elliptic functions as a reference. I’ve skimmed through four chapters and decided that it is probably not very suitable for the complex analysis course later in the year. It’s a pity because the international version of Ahlfors is readily available locally and costs less than US$20.

I’ve been looking at several nice books, Stein and Shakachi, Gamelin and Needham.

All seem pretty promising, especially Needham’s Visual Complex Analysis. I have read many good reviews of the book, and now that I’m starting to look at the contents seriously, I agree it’s very good, and unconventional.

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Oliver Heaviside

Posted by tpc at April 30th, 2008

I have no idea who is Heaviside until I started to teach this course which included Laplace transforms. That got me really interested and I checked out P. Nahin’s biography from the library. Perusing the borrowing slip, the book was last borrowed in Sep 2006, prior to that Mar 1990. 16 long years.

Meanwhile, in the preface a quote attributed to Lazarus Long:

Anyone who cannot cope with mathematics is not fully human. At best he is a tolerable subhuman who has learned to wear shoes, bathe, and not make messes in the house.

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Morrie’s Law

Posted by tpc at April 19th, 2008

Came across this curious identity while preparing some trigo notes.
 \cos (20^\circ) \cos (40^\circ) \cos (80^\circ) =\frac{1}{8}
According to wolfram, it is called Morrie’s Law by Feynman after his childhood friend who showed it to him. Wow, all I need now is some future nobel laureate with a big mouth (no disrespect intended) , so that some cute little identity might be named after me.

It’s easy to prove Morrie, just multiply \sin x to the numerator and denominator for each of the 3 angles. Use the double angle formula for sine and then cancel away with \cos x = \sin (90^\circ -x)

The general formula, easily proven with induction is the following:
\displaystyle 2^n \prod^{n-1}_{k=0} \cos (2^k x) = \frac{\sin(2^n x)}{\sin x}

Posted in Trigonometry| No Comments |