# 2-8.

For each of the following pairs of functions, either $f(n)$ is in $O(g(n))$, $f(n)$ is in $\Omega(g(n))$, or $f(n)=\Theta(g(n))$. Determine which relationship is correct and briefly explain why.

## $f(n)=\log n^2$; $g(n)=\log n$ + $5$

Answer: $\log n^2 = \Theta (\log n + 5)$

Solution:

$\log n^2 = 2 \times \log n$

$2 \times \log n \le 2 \times \log n + 10$

$\log n^2 \le 2 (\log n + 5)$

$\log n^2 \le C (\log n + 5)$ (where $C=2$)

$\log n^2 = O (\log n + 5)$

Also:

$\log n + 5 \le \log n + 5 \log n$

$\log n + 5 \le 6 \log n$

$\log n + 5 \le 3 \times 2 \log n$

$3 \times \log n^2 \ge \log n + 5$

$\log n^2 \ge C \times (\log n + 5)$ (Where $C =\frac{1}{3}$)

$log n^2 = \Omega (\log n + 5)$

And therefore:

$log n^2 = \Theta (\log n + 5)$

## $f(n)=\sqrt{n}$; $g(n)=\log(n^2)$

Answer: $f(n) = \Omega(g(n))$

Solution:

$g(n) = \log (n^2) = 2 * \log (n)$

$\lim_{n \to \infty} \frac{\sqrt{n}}{2 * log(n)} = 2 * \lim_{n \to \infty} \frac{\sqrt{n}}{log(n)} = \infty$

## $f(n)=\log^2(n)$; $g(n)=\log (n)$

Answer: $f(n) = \Omega(g(n))$

Solution:

$\lim_{n \to \infty} \frac{log^2(n)}{log(n)} = \lim_{n \to \infty} log(n) = \infty$

## $f(n)=n$; $g(n)=\log^2(n)$

Answer: $f(n) = \Omega(g(n))$

Solution:

$\lim_{n \to \infty} \frac{n}{\log^2(n)} = \lim_{n \to \infty} ((\frac{\sqrt{n}}{\log(n)})^2) = (\lim_{n \to \infty} \frac{\sqrt{n}}{\log(n)})^2 = \infty$

## $f(n)=n * \log(n) + n$; $g(n)=\log (n)$

Answer: $f(n) = \Omega(g(n))$

Solution:

$\lim_{n \to \infty} \frac{n * \log(n) + n}{log(n)} = \lim_{n \to \infty} (\frac{n * \log(n)}{log(n)} + \frac{n}{log(n)}) = \lim_{n \to \infty} (n + \frac{n}{log(n)}) = \infty$

## $f(n)=10$; $g(n)=\log (10)$

Answer: $f(n) = \Theta(g(n))$

Solution: Both are constants. Constants are always within a constant factor, $c$, of each other (as $n \rightarrow \infty$).

## $f(n)=2^n$; $g(n)=10n^2$

Answer: $f(n) = \Omega(g(n))$

Solution:

$\lim_{n \to \infty} \frac{2^n}{10n^2} = \frac{1}{10} (\lim_{n \to \infty} \frac{2^n}{n^2}) = \frac{1}{10} (\lim_{n \to \infty} \frac{n * 2^{n-1}}{2 * n})$ (L'Hopital's Rule)

$= \frac{1}{10} (\lim_{n \to \infty} 2^{n-2}) = \infty$

## $f(n)=2^n$; $g(n)=3^n$

Answer: $f(n) = O(g(n))$

Solution:

$\lim_{n \to \infty} \frac{2^n}{3^n} = \lim_{n \to \infty} (\frac{2}{3})^n = 0$