Question

Show that \(x\log x\) is \(O({x^2})\) and \({x^2}\)is not \(O(x\log x)\).

Short answer

Hence, we obtain \(x\log x\) is \(O({x^2})\) and \({x^2}\)is not \(O(x\log x)\)

Step-by-step solution

Step 1:

Let \(f(x) = x\log x\) and \(g(x) = {x^2}\)

When \(x > 1\), \(\log x > 0\)

Let \(k = 1\)

\( \Rightarrow C = 1\)

\( \Rightarrow f(x) = x\log x\) is \(O({x^2})\)

Step 2:

Let \(f(x) = {x^2}\) and \(g(x) = x\log x\)

\( \Rightarrow \left| {{x^2}} \right| \le C\left| {x\log x} \right| = \left| {Cx\log x} \right|\)where \(x > k\)

Here, we are getting a contradiction because\(\left| x \right| > \left| {\log x} \right|\)with\(x > 0\)

\( \Rightarrow \)\(f(x) = {x^2}\)is not \(O(x\log x)\)