Question

Prove that if a,b∈R and b>0, then (n+a) ᵇ∈Θ(nᵇ).

Answer 1

To prove that (n+a)^b ∈ Θ(n^b), we need to show that there exist positive constants c1, c2, and n0 such that for all n ≥ n0, c1n^b ≤ (n+a)^b ≤ c2n^b.

Step-by-step explanation:

To prove that (n+a)b ∈ Θ(nb), we need to show that there exist positive constants c1, c2, and n0 such that for all n ≥ n0, c1nb ≤ (n+a)b ≤ c2nb.

For n ≥ 0, we have (n+a)b = nb + bnb-1a + ... + bn2ab-2 + banb-1 + ab. Since a and b are real numbers and b > 0, each term in the expansion is positive.

Therefore, we can choose c1 = 1 and c2 = (1+a)b as the positive constants. For a given n, we have:

c1nb = nb

c2nb = (1+a)bnb = (n+a)b

Hence, (n+a)b ∈ Θ(nb) is proved.