To approach this, we start by writing out the definition of Θ with *f*(*n*) =
4*n*^{3} and *g*(*n*) = *n*^{3}: Show that ∃ *c*_{1}, *c*_{2}, *n*_{0}≥ 0 such
that for all *n* ≥ *n*_{0},

c_{1}n^{3}≤ 4n^{3}≤c_{2}n^{3}.

Try *c*_{1} = *c*_{2} = 4; *n*_{0} = 1:

4n^{3}≤ 4n^{3}≤ 4n^{3}.

That was too easy! It's harder when you need to absorb a lower order term:

Show that ∃ *c*_{1}, *c*_{2}, *n*_{0}≥ 0 such
that ∀ *n* ≥ *n*_{0},

c_{1}n^{3}≤ 4n^{3}+ 2n≤c_{2}n^{3}

Try *c*_{1} = 4; *c*_{2} = 5. (I chose *c*_{1} = 4 to
make the first term more like the second: certainly adding 2*n* will make it larger. I chose 5
since the right hand term has to grow larger.)

4n^{3}≤ 4n^{3}+ 2n≤ 5n^{3}

Divide by *n*^{3}:

4 ≤ 4 + 2/n^{2}≤ 5

This is true for all *n* ≥ *n*_{0} = 2, since the term
2/*n*^{2} will always be larger than 0 (satisfying the first inequality) but less than
1 (satisfying the second inequality).

You should be able to do similar analyses.

Nodari Sitchinava (based on material by Dan Suthers) Last modified: Mon Nov 19 18:23:20 HST 2012