Supplementary answer to Q6 (Tutorial Exercise - Analysis)
A robust method to prove 2n2-5n=(n2)
6. To prove 2n2-5n=(n2), we need to find out positive c1,c2,n0 such that c1*n2  2n2-5n  c2*n2, for
all n  n0.
Dividing by n2 yields: c1  2-5/n  c2
For 2-5/n  c2 Consider that 5/n is positive whenever n1.
=> 2-5/n  2 whenever n1
=> 2-5/n  c2 whenever c22 and nn0, where n01.
=> 2n2-5n  c2*n2 whenever c22 and nn0, where n01.
For c1  2-5/n Consider that (2-5/n) is an increasing function,
and (2-5/n)=0 happens when n=2.5
=> (2-5/n) 0 whenever n2.5
=> (2-5/n) 0 whenever n3 (more happy to see n an integer!!)
But actually when n=3, 2-5/n=1/3.
Since (2-5/n) is an increasing function,
we can write 1/3 2-5/n whenever n3.
Or, c1  2-5/n whenever c11/3 and nn0, where n03.
Or, c1*n2  2n2-5n whenever c11/3 and nn0, where n03.
Question: Can we choose a smaller c1 and smaller n0 to make c1 2-5/n?
Answer: not possible: since c1 has to be positive and we want n to be integer.
Conclusion, “c1*n2  2n2-5n  c2*n2, for all n  n0” is true given c1=1/3, c2=2, n0=3. (Indeed, it is
also true for any c1, c2 and n0 such that they are  1/3,  2, and  3 respectively). Therefore, 2n25n=(n2).