TADM2E 2.8

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 $