3. Let p(n) = 0 ani, where aa > 0, be a degree-d polynomial in n, and let k be a constant. Prove the following properties. if k ≥d, then p(n) = 0(nk); if k ≤d, then p(n) = N(nk); if k = d, then p(n) = 0(nk); if k > d, then p(n) = o(nk); if k
Can you give a detailed answer for part D and E
d. If k > d, then p(n) = o(n^k)
This means that p(n) grows slower than n^k as n approaches infinity. To prove this property, we need to show that there exists a positive constant c such that |p(n)| <= c * |n^k| for all n >= n_0, where n_0 is some positive integer.
Consider the following:
|p(n)| = |Σ l=0^d a_l * n^l| <= |a_d| * |n^d| + |Σ l=0^d-1 a_l * n^l| <= |a_d| * |n^d| + |(a_0 + a_1 * n + a_2 * n^2 + ... + a_d-1 * n^d-1) * n^(d-k)| <= |a_d| * |n^d| + M * |n^(d-k)|, where M is some positive constant
Since k > d, we have d-k < 0, and so n^(d-k) approaches 0 as n approaches infinity. Hence, for all n >= n_0, we have
|p(n)| <= |a_d| * |n^d| + M * |n^(d-k)| <= (|a_d| + M) * |n^d|
Since k > d, we have n^k approaches infinity faster than n^d as n approaches infinity. Hence, we can choose c = |a_d| + M and n_0 such that for all n >= n_0,
|p(n)| <= c * |n^k|
Therefore, p(n) = o(n^k).
Trending now
This is a popular solution!
Step by step
Solved in 2 steps
