Let f(n) = O(n), g(n) = θ(n), and h(n) = Ω(n). Then f(n). h(n) + g(n) is_______________

Question

let us consider f(n) is log(n) and g(n) = n and h(n) = n^2.

since logn<= n

        n2 >= n for all values so given above equality holds true

but when we substitute n+((logn)(n2) = O(n2) but ans is Ω(n) can somebody wxplain this plz

Answer 1

You can also take h(n) = n³, and then the upper bound changes. But whatever you take the lower bound of f(n) won't change and remains n because of the θ(n) term. 

Comments

i found a solution i have some problems regarding that solution

f(n)=O(n), g(n)=Ω(n) , h(n)=Θ(n).

g(n)+h(n).f(n)

= Ω(n) +O(n).Θ(n)

= Ω(n) +[Θ(n) <-->Θ(n²) ] what is the meaning of the [Θ(n) <-->Θ(n²) ] ?

As i think , that means the value of O(n).Θ(n) can vary between Θ(n) to Θ(n²) or we can say like O(n) can be 1 or n. and Θ(n) can be n
 

=Ω(n)

the above equation  Ω(n) +[Θ(n) <-->Θ(n²) ]  results in Ω(n) . how?

is it like we should opt higher function or Max(Ω(n) , [Θ(n) <-->Θ(n²) ] )

and Ω(n) will have higher value than Θ(n) or Θ(n²)  ??
http://math.stackexchange.com/questions/267252/how-to-prove-that-maxfn-gn-thetafn-gn
 

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If we just use the English meaning of the symbols, it becomes very easy. $\Omega$ refers to asymptotically lower or equal. So, $f$ is either asymptotically growing equal as $n$ or is growing faster than $n$- we do not know how much faster. Similarly, $O(n)$ does no say how much slower.

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please check my approach--->

let n=25.

For f(n), n<=25. Take n=18.

For g(n),n=25.Take n=25

For h(n),n>=25.Take n=35

f(n). h(n)+f(n) = 18*35+18=646

646=omega(25) as for f(n) = omega(n) . n can be greater than or equal to n

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U said h(n) can be  n^3 anything .

Now given that h(n)=Omega(n) , but don't we have to chose tightest lower bound . I mean for h(n) to be n^3 , the question writer would have written h(n) = Omega(n^2) . Then The Tightest Could Have been n^3.