Let \(p(n)=a_{k} n^{k}+a_{k-1} n^{k-1}+\ldots+a_{1} n+a_{0},\) where \(a_{k}>0 .\) Using the Properties or order in Section \(1.4 .2,\) show that \(p(n) \in \Theta\left(n^{k}\right)\)

Short Answer

Expert verified
\(p(n) = a_{k}n^{k}+a_{k-1}n^{k-1}+...+a_{1}n+a_{0}\) is in \(\Theta(n^{k})\).

Step by step solution

01

Step 1

We first consider the lower bound \(c_1n^{k}\). The polynomial \(p(n)\) has a nonnegative coefficient \(a_{k}\) for its \(n^{k}\) term. Therefore, for \(n \geq 1\), the inequality \(a_{k}n^{k} \leq p(n)\) holds. Hence, we can choose \(c_1 = a_{k}\) and \(n_{0}=1\) for the lower bound requirement.
02

Step 2

Next, we consider the upper bound \(c_2n^{k}\). For any \(n \geq 1\), every term in \(p(n)\) is less than or equal to \(a_{i}n^{k}\), since \(n^{i} \leq n^{k}\) because \(k \geq i\) and \(n \geq 1\). Summing over all \(k+1\) terms, we can get an upper bound for \(p(n)\) by summing over all \(k+1\) values of \(a_{i}n^{k}\). For each term, we can choose \(c_2\) such that \(c_2 = (k+1)\max_i(a_i)\).
03

Step 3

Since we have identified constants \(c_1, c_2\) and \(n_{0}\) such that for all \(n \geq n_{0}\) the inequality \(c_1n^{k} \leq p(n) \leq c_2n^{k}\) holds, we can conclude that \(p(n) \in \Theta(n^{k})\).

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