CS 173

Spring 2021

Spring 2021

Let's see how to prove one of the relationships in our ordering of primitive functions: \(n^2 << 2^n\).

We need to show that \(\lim_{n \rightarrow \infty} \frac{n^2}{2^n} = 0\).

Recall from calculus that \(a^n\) limits to zero as n gets big, if \(a < 1\). So, in particular, \(\lim_{n \rightarrow \infty} (3/4)^n = 0\).

We're going to show that \(\frac{n^2}{2^n}\) limits to zero by showing that the terms of this sequence are smaller than the terms of \((3/4)^n\), with a bit of an offset in the indexing. Since the larger sequence limits to zero, the smaller one must also do so.

To show this, we need the following lemma:

Lemma: If n is big enough (n >= 5), then \((n+1)^2 < \frac{3}{2} n^2\).

This requires a short direct proof.

To understand how the two sequences are related, let's compare \(\frac{n^2}{2^n}\) to \(\frac{ (n+1)^2}{2^{n+1}}\). The lemma says that the top of the fraction increases by less than \(\frac{3}{2}\). The bottom increases by a factor of 2. So the whole fraction increases by less than \(\frac{3}{4}\). As we move forward in the sequence, we accumulate factors of \(\frac{3}{4}\).

So, if we can find a starting relationship between terms of the two sequences, we can pin down the details of a relationship between the sequences. With a bit of fiddling around plugging in small constants for n, we find a good starting point:

At n=5

\(\frac{n^2}{2^n} = \frac{25}{32}\)

\((3/4)^{n-5} = (3/4)^0 = 1\)

So \(\frac{n^2}{2^n} < (3/4)^{n-5}\) at n=5.

So here's the claim we'll prove by induction:

Claim: \(\frac{n^2}{2^n} < (3/4)^{n-5}\) for all \(n \ge 5\).

Proof by induction on n.

Base case: At n=5, \(\frac{n^2}{2^n} = \frac{25}{32}\). Also, \((3/4)^{n-5} = (3/4)^0 = 1\). So \(\frac{n^2}{2^n} < (3/4)^{n-5}\).

Inductive hypothesis: suppose that \(\frac{n^2}{2^n} < (3/4)^{n-5}\) for n = 5, 6, ..., k.

Rest of inductive step: By the lemma \(\frac{(k+1)^2}{2^{k+1}} < (\frac{3}{2}) \frac{k^2}{2^{k+1}} = (\frac{3}{2}) \frac{k^2}{2\cdot 2^k} = (\frac{3}{4}) \frac{k^2}{2^k} \)

By the inductive hypothesis, \(\frac{k^2}{2^k} < (3/4)^{k-5}\). Substituting into the previous equation gives us: \[\frac{(k+1)^2}{2^{k+1}} < (\frac{3}{4}) \frac{k^2}{2^k} < (\frac{3}{4}) (3/4)^{k-5} = (3/4)^{k-4} \]

So \(\frac{(k+1)^2}{2^{k+1}} < (3/4)^{k-4}\), which is what we needed to prove.

Our final proof was fairly simple. However, we needed to fiddle around with the pieces of the problem to figure out exactly what constants we needed and how to compare bits of algebra. Inequality induction problems often require more scratchwork than problems involving equality.

Suppose you need to prove that \(a < b\), where a and b are two bits of algebra. There's a gap in size between a and b. Frequently you have to hypothesize som intermediate term c, for which \(a < c < b\) in order to make your algebra work. Sometimes you have to fiddle around with scratchwork to find the right c that makes it easy to prove that \(a < c\) and \(c < b\).