# 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.

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.