College Math Teaching

August 7, 2014

Letting complex algebra make our calculus lives easier

Filed under: basic algebra, calculus, complex variables — Tags: , — collegemathteaching @ 1:37 am

If one wants to use complex arithmetic in elementary calculus, one should, of course, verify a few things first. One might talk about elementary complex arithmetic and about complex valued functions of a real variable at an elementary level; e. g. f(x) + ig(x) . Then one might discuss Euler’s formula: e^{ix} = cos(x) + isin(x) and show that the usual laws of differentiation hold; i. e. show that \frac{d}{dx} e^{ix} = ie^{ix} and one might show that (e^{ix})^k = e^{ikx} for k an integer. The latter involves some dreary trigonometry but, by doing this ONCE at the outset, one is spared of having to repeat it later.

This is what I mean: suppose we encounter cos^n(x) where n is an even integer. I use an even integer power because \int cos^n(x) dx is more challenging to evaluate when n is even.

Coming up with the general formula can be left as an exercise in using the binomial theorem. But I’ll demonstrate what is going on when, say, n = 8 .

cos^8(x) = (\frac{e^{ix} + e^{-ix}}{2})^8 =

\frac{1}{2^8} (e^{i8x} + 8 e^{i7x}e^{-ix} + 28 e^{i6x}e^{-i2x} + 56 e^{i5x}e^{-i3x} + 70e^{i4x}e^{-i4x} + 56 e^{i3x}e^{-i5x} + 28e^{i2x}e^{-i6x} + 8 e^{ix}e^{-i7x} + e^{-i8x})

= \frac{1}{2^8}((e^{i8x}+e^{-i8x}) + 8(e^{i6x}+e^{-i6x}) + 28(e^{i4x}+e^{-i4x})+  56(e^{i2x}+e^{-i2x})+ 70) =

\frac{70}{2^8} + \frac{1}{2^7}(cos(8x) + 8cos(6x) + 28cos(4x) +56cos(2x))

So it follows reasonably easily that, for n even,

cos^n(x)  = \frac{1}{2^{n-1}}\Sigma^{\frac{n}{2}-1}_{k=0} (\binom{n}{k}cos((n-2k)x)+\frac{\binom{n}{\frac{n}{2}}}{2^n}

So integration should be a breeze. Lets see about things like, say,

cos(kx)sin(nx) = \frac{1}{(2)(2i)} (e^{ikx}+e^{-ikx})(e^{inx}-ie^{-inx}) =

\frac{1}{4i}((e^{i(k+n)x} - e^{-i(k+n)x}) + (e^{i(n-k)x}-e^{-i(n-k)x}) = \frac{1}{2}(sin((k+n)x) + sin((n-k)x)

Of course these are known formulas, but their derivation is relatively simple when one uses complex expressions.


August 6, 2014

Where “j” comes from

I laughed at what was said from 30:30 to 31:05 or so:

If you are wondering why your engineering students want to use j = \sqrt{-1} is is because, in electrical engineering, i usually stands for “current”.

Though many of you know this, this lesson also gives an excellent reason to use the complex form of the Fourier series; e. g. if f is piece wise smooth and has period 1, write f(x) = \Sigma^{k = \infty}_{k=-\infty}c_k e^{i 2k\pi x} (usual abuse of the equals sign) rather than writing it out in sines and cosines. of course, \overline{c_{-k}} = c_k if f is real valued.

How is this easier? Well, when you give a demonstration as to what the coefficients have to be (assuming that the series exists to begin with, the orthogonality condition is very easy to deal with. Calculate: c_m= \int^1_0 e^{i 2k\pi t}e^{i 2m\pi x} dx for when k \ne m . There is nothing to it; easy integral. Of course, one has to demonstrate the validity of e^{ix} = cos(x) + isin(x) and show that the usual differentiation rules work ahead of time, but you need to do that only once.

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