# Finding constants C and k for big-O of fraction of polynomials

I am a teaching assistant on a course for computer science students where we recently talked about big-O notation. For this course I would like to teach the students a general method for finding the constants $C$ and $k$ in the definition of big-O $$|f(x)| \leq C|g(x)|, x > k,$$ when the function $f(x)$ is a fraction of polynomials.

I have tried to concoct my own method, but I am unsure of its correctness. I was inspired by the easy method of finding $C$ and $k$ for a polynomial where for example we can show that $x^3+2x+3$ is $O(x^3)$ by $$x^3+2x+3 \leq x^3+2x^3+3x^3 = 6x^3$$ to find $C = 6$ and $k = 1$.

Now, for a fraction of polynomials I am unsure what to do with the denominator. My attempt at a general method is as follows. I would like to show that $\frac{x^4+x^2x1}{x^3+1}$ is $O(x)$. First I divide by the highest term in the denominator to get a $1$ in the denominator: $$\frac{(x^4+x^2+1)/x^3}{(x^3+1)/x^3} = \frac{x+\frac{1}{x}+\frac{1}{x^2}}{1+\frac{1}{x^3}}$$ Now I argue, somewhat analogously to the previous example, that the fractions in the numerator must be less than (when x > 0) x, and since a smaller denominator makes the expression smaller, setting all the terms in denominator except the $1$ to $0$, I obtain the inequality $$\frac{x+\frac{1}{x}+\frac{1}{x^2}}{1+\frac{1}{x^3}} \leq \frac{x+x+x}{1+0} = 3x$$ and I find $C = 3$ and $k = 1$.

Now my question is, does this homebrewed method actually work or is it complete nonsense? And if it is nonsense, does anybody know of another general method for finding $C$ and $k$ in instances like this?

Note that I need to find the constants $C$ and $k$, not just show that a given function is big-O of some other function, and the students have had no course in calculus, so I can use no concepts from there, such as limits.

• By the way, your "fractions of polynomials" are known as rational functions. – David Richerby Feb 28 '14 at 22:19

Remember that $O(-)$ is just an upper bound.

Given a rational function $p(x)/q(x)$, you already know how to find $c$, $k$ and $x_0$ such that $|p(x)|\leq c|x^k|$ for $x>x_0$. By similar arguments, you can show that $|q(x)|\ge 1$ for all $x$ greater than some $x_1$. Therefore, for all $x>\max\{x_0,x_1\}$, we have $|p(x)/q(x)| \leq |p(x)| \leq c|x^k|$ so $p(x)/q(x) = O(x^k)$.

• Of course, this bound is not particularly tight in general... – Yuval Filmus Mar 1 '14 at 1:52
• Agreed but the question doesn't ask for tight bounds: it just asks for a method of finding the two constants. And it's not usually worth spending a whole lot of time quantifying exactly what "for large enough $n$" means. Normally, it's sufficient to exhibit some value of the constants that works and optimizing them doesn't gain you anything. – David Richerby Mar 1 '14 at 10:50
• This is a very simple and convincing argument, I'll try to use it. Thanks! – mrp Mar 2 '14 at 8:03

It works if the denominator doesn't have negative coefficients. Consider, however, $$\frac{x}{x-1}.$$ As $x$ tends to infinity, this is $O(1)$. However, your method would give $$\frac{x}{x-1} = \frac{1}{1-\tfrac{1}{x}} \leq \frac{1}{1-1} = \infty,$$ which is not very helpful. The problem is that when $x = 1$, the denominator is not positive. In this case it is zero, but for an example like $$\frac{x}{x-2} = \frac{1}{1-\tfrac{2}{x}} \leq \frac{1}{-1} = -1,$$ the inequality is in fact wrong.

Often when considering $O(f(x))$, we assume that $f(x)$ is always positive. But sometimes it isn't, and then the assumption that we're making is that $f(x)$ is eventually positive. This is not usually explained when big O notation is defined, but it's the reason behind the extra quantifier "for large enough $x$" which appears in the full definition of big O; this quantifier is only needed when $f(x)$ is not always positive but only eventually positive.

The way to fix the argument is now clear. If $f(x)$ is eventually positive, then take some $y$ such that $f(x) > 0$ for all $y \geq x$, and now apply the method. For example, if $f(x) = x/(x-1)$ then any $y > 1$ would do, say $y = 2$. In that case we have $$f(x) = \frac{1}{1-\tfrac{1}{x}} \leq \frac{1}{1-\frac{1}{2}} = 2$$ for $x \geq 2$.

More generally, suppose that $f(x) = C p(x)/q(x)$, where $p,q$ are monic polynomials (leading coefficient is $1$) and are eventually positive, and $C > 0$. Then for all $\epsilon > 0$ there is $y$ such that for all $x \geq y$, $$(C-\epsilon) x^{\deg p - \deg q} < f(x) < (C+\epsilon) x^{\deg p - \deg q}.$$ (Exercise.)

Compare the definition of the Bachman-Landau notations to the definition of a limit when $n \to \infty$, the relation between them is enough for cars like this.

• From the question: "the students have had no course in calculus, so I can use no concepts from there, such as limits." – David Richerby Mar 1 '14 at 16:05