Discussion
Given the following equalities : \(E_{1} : n^{K+\epsilon} + n^{K} \lg n = \theta(n^{K+\epsilon})\) for all fixed K and \(\epsilon\), \(K \ge 0\) and \(\epsilon \gt 0\). \(E_{2} : n^{3} 2^{n} + 6 n^{2} 3^{n} = O(n^{3} 2^{n})\). Which of the following is true ?
Answer: B
\(E_{1}\) is correct and \(E_{2}\) is not correct.
For \(E_{1}\), \(n^{K} \lg n\) is of lower order than \(n^{K+\epsilon}\) for every \(\epsilon \gt 0\), so the sum is \(\Theta(n^{K+\epsilon})\). For \(E_{2}\), \(3^{n}\) grows faster than \(2^{n}\) by an exponential factor, so \(6n^{2}3^{n}\) is not \(O(n^{3}2^{n})\).
So options that call \(E_{1}\) false or \(E_{2}\) true are wrong.
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