Question

please be clear with the steps taken and understandable

1. Prove that if f(n) = Θ(n2) for all f(n), then ΣΑ(n)-6(n3). i=1 2. Prove that if f.(n) are linear functions - i.e., that f(n)-Θ(n) for all Tn A(n) then Σ if.(n) = Θ(n3). Y definition of Big-Oh. ou are not required to use the formal i1

Answer 1

1.

Since f_i(n) = Θ(n²), this means

0 <= c1.n² <= f_i(n) <= c2.n² where c1 and c2 are positive constants.

Multiplying this equation with n, we get

0 <= c1.n² * n <= n * f_i(n) <= c2.n² * n

or, 0 <= c1.n³ <= <= c2.n³

>fi(n)

Hence, = Θ(n³) (proved)

>fi(n)

2.

Given that f_i(n) = Θ(n), this means that

0 <= c1*n <= f_i(n) <= c2*n where c1 and c2 are positive constants.

Let f_i(i) = c*n where c1<= c <= c2

Now, ∑ i*f_i(n) = (1 + 2 + 3 + ..... + n) * c * n = c * n * (n² + n)/2 = c * (n³/2 + n²/2)

Clearly, for c1 = c and c2 = 2c and n >= 0, we have

0 <= c1*n³ <= c * (n³/2 + n²/2) <= c3 * n³

Hence, ∑ i*f_i(n = Θ(n³)