### And Into The Black

If somebody comes up to you out of the blue and says, okay, a Rational Function has the form $R(x) = {{f(x)}\over{g(x)}}$, where f and g are polynomial functions, why then, there are any number of things about this definition that you might want to know next. Hey Vlorbik (you might say for example), how about giving us an example for hecksake? Outstanding.

“Show me one that is; show me one that isn’t”—if there could be such a thing as training a subject in straight thinking, maybe this would then be drilled into students (like the economists’ excuses for power’s abuses): find the key examples! (And [so-called] “counterexamples”; entire [useful, fascinating] books have been devoted to these.)

But, then, what makes an example “key” (this line of investigation might continue)? And if I had to pick one place to look first for an answer, I might quickly settle on the “simplest” examples. In considering the definition of Rational Functions (RFs), the well-prepared mind soon finds itself being drawn to consideration of, not the simplest RF—that would be the constant at zero—but rather the simplest RF that’s not a polynomial (the point here is that since polynomials are used in the definition we’re considering here, we’re already assuming that some theory of polynomials is [at least partly] in place; we’re looking for something “new”).

So. That would be $[x\mapsto {1\over x}]$, the justly-famous reciprocal function. How so? (Thanks for asking.) Well, I’ve already ruled out the zero function as too simple; so the numerator (also known as f—part of my job is to help people become at least a little more comfortable about simultaneous consideration of two different points of view for one same phenomenon…) also can’t be the zero function. Let’s see. The next simplest thing would be f(x) = 1 (the constant at 1). What about downstairs? Well g oughtn’t to be also a constant (since then R itself would be a constant… which is a polynomial [of degree zero]… and so has already been dismissed from consideration]); what’s the next simplest thing after that? Well, not degree zero… degree one, then. Who’s got degree one? Things with x to the first power, right? What’s the simplest thing with x to the first power in it? The question answers itself: x itself.

There it is. One has just stared the problem down: look straight at it until it tells you something. R(x) = 1/x sure enough must be the key example here: the simplest rational function that’s not a polynomial. The awesome simplicity of this gorgeous curve is even more striking (to my eyes) when considered as the graph of xy = 1 (rather than y = 1/x—fractions are always hard).

Before going on to start playing with the reciprocal function—by applying the Transformations considered in the first part of our course—let me mention here that the “curve” in question is a hyperbola. This seems not to be very well-known (by contrast, many even among the doomed—whose miserable lot it is to take “remedial” math classes fruitlessly because they’ve convinced themselves before even getting started that it will never make sense—consider it common knowlege that the graph of y = x^2 is a “parabola”); many a veteran of Math 150 will know something (for at least a few weeks) about hyperbolas of the considerably more complicated form ${{(x-h)^2}\over{a^2}} - {{(y-k)^2}\over{b^2}} = 1$ but with no inkling that the graph of the reciprocal function is also hyperbolic. This somewhat depressing state of affairs seems to be a result of not having asked and answered the “key example” question often enough when introducing Analytic Geometry.

Anyhow, now we hit it with the shifts-and-scalings from week one; the result is the (rich and strange) set of linear fractional transformations:
$\mu (x) = {{Ax + B}\over {Cx + D}}$
(“quotients of linear functions”; one insists here that AD – BC is nonzero [exercise: find out why]). There are, anyway, entire Chapters of books about these. Somebody’s probably done an entire course in ’em. Certainly somebody could. But not me. This is the kind of thing that sometimes makes me almost sorry I didn’t try harder to get back into the Pros.