Show that for any real constants \(a\) and \(b\), where \(b > 0\),

\[(n + a)^b =\Theta(n^b)\]

Note that, \(n + a \le 2n\), when \(\vert a \vert \le n\)

Also note, \(n + a \ge n/2\), when \(\vert a \vert \le n/2\)

Therefore, when \(n \ge 2 \vert a \vert\), both of the above inequalities hold true, and we can write,

\[0 \le \frac n 2 \le n + a \le 2n\]As \(b > 0\), we can raise all the terms of the previous inequality to the power of \(b\) without breaking the inequality:

\[\begin {aligned} 0 \le (\frac n 2)^b & \le (n + a)^b \le (2n)^b \\ 0 \le \frac 1 {2^b}n^b & \le (n + a)^b \le 2^bn^b \end {aligned}\]So, \((n + a)^b = \Theta(n^b)\) because there exists \(c_1 = 1/{2^b}\), \(c_2 = 2^b\), and \(n_0 = 2 \vert a \vert\).