An Example with Proof

$6 n \log n + \sqrt{n} \log^2 n = \Theta(n \log n)$.

Need to find n0, c1, and c2, such that

$c_1 n \log n \le 6 n \log n + \sqrt{n} \log^2 n \le c_2 n \log n$

c1 = 5, c2 = 7, n0 = ...

Note: Implications go up. That is, we want to find something at the bottom that will logically imply the statement we're starting with.

\begin{eqnarraystar}
5 n_0 \log n_0 &\le& 6 n_0 \log n_0 + \sqrt{n_0} \log^2 n_0...
 ... n_0 \gt 1 \\  0 &\le& n_0 \log n_0, n_0 \gt 1 \\  n_0 &\gt& 1\end{eqnarraystar}

\begin{eqnarraystar}
6 n_0 \log n_0 + \sqrt{n_0} \log^2 n_0 &\le& 7 n_0 \log n_0...
 ...n_0 &\le& n_0 \  \log n_0 &\le& \sqrt{n_0} \ n_0 &\gt& 0 \ \end{eqnarraystar}

So, n0 = 3 suffices.


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