# Polynomials

I had an archimedes moment this morning just after I woke up. I was thinking about the function tan-1(x-1) and it’s peculiar discontinuity. I wondered if you could do that kind of “finite discontinuity” with a composition of ratios of polynomials, and concluded that you can iff the limits at positive or negative infinity of the ratio are noninfinite and not identical.

Theorem: For any two polynomials P and Q, let f(x) = P(x)/Q(x). Then (limx -> inf f(x) = +-inf, and limx->-inff(x) = +-inf) or limx->inff(x) = limx->-inff(x).

Proof: (Informal) If P and Q are of different degrees, then both limits are +-infinity or zero. If P and Q are both degree zero, then both limits are a constant, namely P(0)/Q(0). Assume that the conjecture is true for degree n polynomials. If P and Q are both degree n+1, then due to L’Hôpital’s rule, the limits are the same as in P'(x)/Q'(x), which are degree n. QED.

So the result is that no, you cannot. If there’s a discontinuity in a composition of ratios of polynomials, it is an infinite one.

About these ads

## 2 thoughts on “Polynomials”

1. Hello… …I was wondering about this “finite discontinuity” you mentioned and how it was you related that to limits at +-infinity? For example, the function f(x) = (x2+2x+1)/(x+1) has a discontinuity at -1 that does not correspond to it’s pulling to +-infinity at that point (even though it’s limits at +-infinity are both infinite and not identical). So, while this doesn’t disprove your theorem (quite nice proof by induction, by the way), I wonder about your conjecture as a whole (though, maybe I am missing the formal meaning of “finite discontinuity”). Care to share?

2. Vote, vote, vote….., be the first to vote my songs at my stuff.
Yeah, thank. :-)