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| − | =2-8.=
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| − | For each of the following pairs of functions, either <math>f(n)</math> is in <math>O(g(n))</math>, <math>f(n)</math> is in <math>\Omega(g(n))</math>, or <math>f(n)=\Theta(g(n))</math>. Determine which relationship is correct and briefly explain why.
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| − | ==<math>f(n)=\log n^2</math>; <math>g(n)=\log n</math> + <math>5</math>==
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| − | Answer: <math>\log n^2 = \Theta (\log n + 5)</math>
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| − | Solution:
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| − | <math>\log n^2 = 2 \times \log n</math>
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| − | <math>2 \times \log n \le 2 \times \log n + 10</math>
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| − | <math>\log n^2 \le 2 (\log n + 5)</math>
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| − | <math>\log n^2 \le C (\log n + 5)</math> (where <math>C=2</math>)
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| − | <math>\log n^2 = O (\log n + 5)</math>
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| − | Also:
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| − | <math>\log n + 5 \le \log n + 5 \log n</math>
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| − | <math>\log n + 5 \le 6 \log n</math>
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| − | <math>\log n + 5 \le 3 \times 2 \log n</math>
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| − | <math>3 \times \log n^2 \ge \log n + 5 </math>
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| − | <math>\log n^2 \ge C \times (\log n + 5)</math> (Where <math>C =\frac{1}{3}</math>)
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| − | <math>log n^2 = \Omega (\log n + 5)</math>
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| − | And therefore:
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| − | <math>log n^2 = \Theta (\log n + 5)</math>
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| − | ==<math>f(n)=\sqrt{n}</math>; <math>g(n)=\log(n^2)</math>==
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| − | Answer: <math>f(n) = \Omega(g(n))</math>
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| − | Solution:
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| − | <math>g(n) = \log (n^2) = 2 * \log (n)</math>
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| − | <math>\lim_{n \to \infty} \frac{\sqrt{n}}{2 * log(n)} = 2 * \lim_{n \to \infty} \frac{\sqrt{n}}{log(n)} = \infty</math>
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| − | ==<math>f(n)=\log^2(n)</math>; <math>g(n)=\log (n)</math>==
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| − | Answer: <math>f(n) = \Omega(g(n))</math>
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| − | Solution:
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| − | <math>\lim_{n \to \infty} \frac{log^2(n)}{log(n)} = \lim_{n \to \infty} log(n) = \infty</math>
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| − | ==<math>f(n)=n</math>; <math>g(n)=\log^2(n)</math>==
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| − | Answer: <math>f(n) = \Omega(g(n))</math>
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| − | Solution:
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| − | <math>\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</math>
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| − | ==<math>f(n)=n * \log(n) + n</math>; <math>g(n)=\log (n)</math>==
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| − | Answer: <math>f(n) = \Omega(g(n))</math>
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| − | Solution:
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| − | <math>\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</math>
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| − | ==<math>f(n)=10</math>; <math>g(n)=\log (10)</math>==
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| − | Answer: <math>f(n) = \Theta(g(n))</math>
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| − | Solution: Both are constants. Constants are always within a constant factor, <math>c</math>, of each other (as <math>n \rightarrow \infty</math>).
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| − | ==<math>f(n)=2^n</math>; <math>g(n)=10n^2</math>==
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| − | Answer: <math>f(n) = \Omega(g(n))</math>
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| − | Solution:
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| − | <math>
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| − | \begin{align}
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| − | &\lim_{n \to \infty} \frac{2^n}{10n^2} \\
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| − | &\implies \frac{1}{10} \left(\lim_{n \to \infty} \frac{2^n}{n^2}\right) \\
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| − | &\implies \frac{1}{10} \left(\lim_{n \to \infty} \frac{\ln(2)\cdot 2^n}{2 \cdot n}\right) \text{L'Hopital's Rule} \\
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| − | &\implies \frac{1}{10} \left(\lim_{n \to \infty} \frac{2\ln(2)\cdot 2^n}{2}\right) \text{L'Hopital's Rule} \\
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| − | &\implies \frac{\ln(2)}{10} \lim_{n \to \infty} 2^n \\
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| − | &\implies \frac{\ln(2)}{10} \lim_{n \to \infty} 2^n = \infty
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| − | \end{align}
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| − | </math>
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| − | ==<math>f(n)=2^n</math>; <math>g(n)=3^n</math>==
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| − | Answer: <math>f(n) = O(g(n))</math>
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| − | Solution:
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| − | <math>\lim_{n \to \infty} \frac{2^n}{3^n} = \lim_{n \to \infty} (\frac{2}{3})^n = 0</math>
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