Self Doubt

Question

Kindly verify these and provide answer to others-

$\\\Theta (n).O(n) = ?\\ \Theta (n).\Omega (n) = \Omega (n)\\ O(n).\Omega (n) = \Omega (n)\text{when function is non decreasing and what if it is decreasing?}\\ \Omega (n)+\Theta (n) =\Omega (n)\\ O(n)+\Theta (n) = ?\\ O(n)+\Omega (n)=\Theta (n)$

Comments

I don't think we can conclude that it will be Θ(n). a simple interpretation is  x>5 and y=5 then x+y should be greater than 5. Right? 

you are taking integers here, but take in terms of n, then you will get what i want to say !

but don't forget, $n^2 + n + 1$ = Θ(n)

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check this -https://gateoverflow.in/11232/let-f-n-%CF%89-n-g-n-o-n-and-h-n-%D1%B3-n?show=11234#a11234

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@Shubhgupta

ya, i did wrong, Thanks for correcting me !

i assumed f(n) = O(n) means f(n) ≥ n, but actually it is f(n) ≤ n.

But the process is same, like

Θ(n).O(n)  = (exactly n) . (≥n) = ≥n$^2$

===> O(n$^2$)

Θ(n).O(n)  = (exactly n) . (≤n) that is equal to  ≤ n$^2$

===> O(n$^2$)

 

Ω(n)+Θ(n)= ( ≥ n ) + ( exactly n ) = ( minimum n ) = Ω(n)