如何评估以下包含渐近符号的表达式？

Question

If

f(n)=ϴ(n),g(n)=ϴ(n) 

and

h(n)=Ω(n) 

Then how to evaluate f(n)g(n)+h(n) ?

I approached like f(n)g(n)=ϴ(n^2) , now what will be Ω(n)+ϴ(n^2) . According to me the lower bound of this expression should be Ω(n) , and upper bound should be O(n^2) , but what should be the tightest bound for this expression?

Answer 1

For some constants k1, k2, l1, l2 and m > 0 , we have:

f(n) is ϴ(n)

    => k1*n < f(n) < k2*n, for n sufficiently large

g(n) is ϴ(n)

    => l1*n < g(n) < g2*n, for n sufficiently large

h(n) is Ω(n)

    => m*n < h(n), for n sufficiently large

Then, f(n)*h(n) :

for f(n) * h(n): 

    k1*l1*n^2 < f(n)*g(n) < k2*l2*n^2, for n sufficiently large

So we can just write p(n) = f(n)*g(n) and use constants c1=k1*l1 and c2=k2*l2 , and we have:

p(n) (= f(n)*g(n)) is in ϴ(n^2), since

    c1*n^2 < p(n) < c2*n^2

Then, finally, what complexity does p(n) + h(n) have? We have:

c1*n^2 + m*n < p(n) + h(n), for n sufficiently large

Since we never got an upper bound on h(n) , we can't really say anything regarding the upper bound on p(n) + h(n) . This is imperative: h(n) in Ω(n) only says that h(n) grows at least as fast as n (linear) asymptotically, but we don't know if this is a tight lower bound. It might be a very sloppy lower bound for a exponential time function.

Subsequently, we can only state something regarding the lower bound:

p(n) + h(n) = f(n)*g(n) + h(n) is in Ω(n^2)

Ie, f(n)*g(n) + h(n) grows at least as n^2 (ie, in Ω(n^2) ) asymptotically.

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A note as to your approach: you are right (as shown above) that f(n)g(n) is in ϴ(n^2) , but note that this implies that a tight lower bound of f(n)g(n) + h(n) can never be less than k*n^2 : ie, f(n)g(n) + h(n) in Ω(n^2) is a given, and a better (tigher) lower bound than what your suggested; Ω(n) . Recall that the fastest growing terms dominate asymptotic behavior.

For reference, see eg

• https://www.khanacademy.org/computing/computer-science/algorithms/asymptotic-notation/a/asymptotic-notation