寻找函数的上界

Question

Example-3 Find upper bound for f(n) = n^4 + 100n^2 + 50

Solution: n^4 + 100n^2 + 50 ≤ 2n^4, for all n ≥ 11 ∴ n^4 + 100n^2 + 50 = O(n^4 ) with c = 2 and n0 = 11

In the above question the solution says n>11 and n-nought is 11. can anybody explain why is it 11? for reference - this is a problem from the Data Structures and Algorithms Made Easy by Narasimha Karumanchi

Answer 1

It doesn't say that n>11 it says that n4 + 100n2 + 50 ≤ 2n4 , for all n ≥ 11 .

Is it true? You can substitute n for 11 in the formula and check it yourself.

How was 11 obtained? By solving the inequality.

Answer 2

It is not finding an upper bound for a function. It is an asymptotic analysis of a function with big-O notation. Hence, the constant c = 11 does not matter for the analysis, and if you can show the inequality is valid for all n greater than any constant, for instance c = 100 , that will be accepted. By the way, you can show that it is true for all n > 11 by the mathematical induction.

Answer 3

f(n) = n^4 + 100n^2 + 50

Intuitively, n^4 grows very fast; n^2 grows less fast than n^4 ; and 50 doesn't grow at all.

However, for small values of n , n^4 < 50 ; additionally, the n^2 term has a factor 100 in front of it. Because of this factor, for small values of n , n^4 < 100 n^2.

But because we have the intuition that n^4 grows much faster than n^2 , we expect that, for n big enough, 100 n^2 + 50 < n^4 .

In order to assert and prove this claim, we need to be more precise on what "for n big enough" means. Your textbook found an exact value; and they claimed: for n ≥ 11, 100 n^2 + 50 < n^4 .

How did they find that? Maybe they solved the inequality for n . Or maybe they just intuited it by noticing that:

100 n^2 = 10 * 10 * n * n`
    n^4 =  n * n  * n * n

Thus n^4 is going to be the bigger of the two as soon as n is bigger than 10.

In conclusion: as soon as n ≥ 11, f(n) < 2 n^4 . Thus, f(n) satisfies the textbook definition for f(n) = O(n^4) .