# Sorting functions by asymptotic growth

Assume I have a list of functions, for example

$\qquad n^{\log \log(n)}, 2^n, n!, n^3, n \ln n, \dots$

How do I sort them asymptotically, i.e. after the relation defined by

$\qquad f \leq_O g \iff f \in O(g)$,

assuming they are indeed pairwise comparable (see also here)? Using the definition of $O$ seems awkward, and it is often hard to prove the existence of suitable constants $c$ and $n_0$.

This is about measures of complexity, so we’re interested in asymptotic behavior as $n \to +\infty$, and we assume that all the functions take only non-negative values ($\forall n, f(n) \ge 0$).

If you want rigorous proof, the following lemma is often useful resp. more handy than the definitions.

If $$c=limc = \lim_{n\to\infty} \frac{f(n)}{g(n)}$$ exists, then

• $$c=0 \qquad \ \,\iff f \in o(g)c=0 \qquad \ \,\iff f \in o(g)$$,
• $$c \in (0,\infty) \iff f \in \Theta(g)c \in (0,\infty) \iff f \in \Theta(g)$$ and
• $$c=\infty \quad \ \ \ \iff f \in \omega(g)c=\infty \quad \ \ \ \iff f \in \omega(g)$$.

With this, you should be able to order most of the functions coming up in algorithm analysis¹. As an exercise, prove it!

Of course you have to be able to calculate the limits accordingly. Some useful tricks to break complicated functions down to basic ones are:

• Express both functions as $$e^{\dots}e^{\dots}$$ and compare the exponents; if their ratio tends to $$00$$ or $$\infty\infty$$, so does the original quotient.

• More generally: if you have a convex, continuously differentiable and strictly increasing function $$hh$$ so that you can re-write your quotient as

$$\qquad \displaystyle \frac{f(n)}{g(n)} = \frac{h(f^*(n))}{h(g^*(n))}\qquad \displaystyle \frac{f(n)}{g(n)} = \frac{h(f^*(n))}{h(g^*(n))}$$,

with $$g^* \in \Omega(1)g^* \in \Omega(1)$$ and

$$\qquad \displaystyle \lim_{n \to \infty} \frac{f^*(n)}{g^*(n)} = \infty\qquad \displaystyle \lim_{n \to \infty} \frac{f^*(n)}{g^*(n)} = \infty$$,

then

$$\qquad \displaystyle \lim_{n \to \infty} \frac{f(n)}{g(n)} = \infty\qquad \displaystyle \lim_{n \to \infty} \frac{f(n)}{g(n)} = \infty$$.

See here for a rigorous proof of this rule (in German).

• Consider continuations of your functions over the reals. You can now use L’Hôpital’s rule; be mindful of its conditions²!

• Have a look at the discrete equivalent, Stolz–Cesàro.

• When factorials pop up, use Stirling’s formula:

$$\qquad \displaystyle n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n\qquad \displaystyle n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n$$

It is also useful to keep a pool of basic relations you prove once and use often, such as:

• logarithms grow slower than polynomials, i.e.

$$\qquad\displaystyle (\log n)^\alpha \in o(n^\beta)\qquad\displaystyle (\log n)^\alpha \in o(n^\beta)$$ for all $$\alpha, \beta > 0\alpha, \beta > 0$$.

• order of polynomials:

$$\qquad\displaystyle n^\alpha \in o(n^\beta)\qquad\displaystyle n^\alpha \in o(n^\beta)$$ for all $$\alpha < \beta\alpha < \beta$$.

• polynomials grow slower than exponentials:

$$\qquad\displaystyle n^\alpha \in o(c^n)\qquad\displaystyle n^\alpha \in o(c^n)$$ for all $$\alpha\alpha$$ and $$c > 1c > 1$$.

It can happen that above lemma is not applicable because the limit does not exist (e.g. when functions oscillate). In this case, consider the following characterisation of Landau classes using limes superior/inferior:

With $$c_s := \limsup_{n \to \infty} \frac{f(n)}{g(n)}c_s := \limsup_{n \to \infty} \frac{f(n)}{g(n)}$$ we have

• $$0 \leq c_s < \infty \iff f \in O(g)0 \leq c_s < \infty \iff f \in O(g)$$ and
• $$c_s = 0 \iff f \in o(g)c_s = 0 \iff f \in o(g)$$.

With $$c_i := \liminf_{n \to \infty} \frac{f(n)}{g(n)}c_i := \liminf_{n \to \infty} \frac{f(n)}{g(n)}$$ we have

• $$0 < c_i \leq \infty \iff f \in \Omega(g)0 < c_i \leq \infty \iff f \in \Omega(g)$$ and
• $$c_i = \infty \iff f \in \omega(g)c_i = \infty \iff f \in \omega(g)$$.

Furthermore,

• $$0 < c_i,c_s < \infty \iff f \in \Theta(g) \iff g \in \Theta(f)0 < c_i,c_s < \infty \iff f \in \Theta(g) \iff g \in \Theta(f)$$ and
• $$c_i = c_s = 1 \iff f \sim g c_i = c_s = 1 \iff f \sim g$$.

Check here and here if you are confused by my notation.

¹ Nota bene: My colleague wrote a Mathematica function that does this successfully for many functions, so the lemma really reduces the task to mechanical computation.