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Problem: Show that for positive integer values of x and y,

$$Θ(x^3 + xy + y^2) = Θ(x^3 + y^2)$$

Solution:

For this specific example, I can show that $xy$ is asymptotically dominated by the other terms by checking two different cases:

$$x >y \implies xy < x^3$$

$$x \le y \implies xy \le y^2$$

Questions:

Is it safe to say that this case by case approach works in general for multivariable asymptotic families? Any relevant theorems?

Are there other ways of proving this? I am especially interested in a limit definition, analogous to the single variable one.

Otay
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  • What bounds are on $x$ and $y$? – Antonio Vargas Feb 12 '17 at 12:38
  • I sort of made this problem up, but I've never heard of setting bounds while looking at asymptotic problems like this. Can you fill me in on what I'm conceptually missing here? – Otay Feb 12 '17 at 23:08
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    If you don't at least make the restriction that $x > 0$ then the statement is trivially false since $(x^3+xy+y^2)/(x^3+y^2)$ is unbounded for arbitrarily large values of $x$ and $y$ near the curve $x^3 + y^2 = 0$.. – Antonio Vargas Feb 13 '17 at 09:20
  • Ahhh thats interesting. I'll change the problem to specify positive values. Thank you very much. – Otay Feb 13 '17 at 09:53
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    There also needs to be some kind of "asymptotic" regime for this statement to be true. For example, are you interested in the truth of the statement as $x,y \to \infty$ in some way? If you're asking whether the statement is true for all $x,y > 0$ then it is still false near $(x,y) = (0,0)$ since $$\left. \frac{x^3 + xy + y^2}{x^3+y^2} \right|_{x = 2^{-1/3}y^{2/3}} \to \infty \qquad \text{as } y \to 0^+$$ – Antonio Vargas Feb 13 '17 at 11:23
  • I would never have thought of these specifications, thank you for being so rigorous with me I appreciate it. I've only ever used asymptotic notation for algorithm complexity, so I think of x and y as integers usually. I'll edit that to the problem. – Otay Feb 14 '17 at 01:09

1 Answers1

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Multivariable asymptotics can be defined in terms of limits involving the infinity norm, which is defined as follows:

$$\Vert x \Vert_{\infty} = \max_i x_i$$

Then for $x \in \mathbb{R}^n$, $f(x) = O(g(x))$ if there exists $c > 0$ such that

$$\lim_{\Vert x \Vert_{\infty} \rightarrow \infty} \left( c g(x) - f(x) \right) \geq 0.$$

where we assume that $f(x), g(x) \geq 0$ are continuous functions.

Equivalently, this means that for every sequence $(a_i)_{i = 1}^{\infty}$ with each $a_i \in \mathbb{R}^n$, where the entry in some coordinate becomes arbitrarily large (i.e. $\lim_{i \rightarrow \infty} \max_j \vert a_{i,j} \vert = \infty$), there exists $c > 0$ such that $f(a_i) \leq c g(a_i)$ for sufficiently large $i$.

It's not clear to me that your case by case analysis works in general. However, your argument does work with the above definition if you apply it to the elements of an arbitrary sequence $(a_i)_{i = 1}^{\infty}$ such that $\lim_{i \rightarrow \infty} \max_j \vert a_{i,j} \vert = \infty$.

You may also find the wikipedia article useful. It has a section on multivariable asymptotics. The definition is more general, but it is not in terms of an explicit limit (this does not appear to be possible in general).

Qudit
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