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Q1) f(n) = n^(1-sin n)

       g(n) = n^(1+cos n)

Relation between them ?

 

Q2) f(n) = n^(1-sin n)

       g(n) = n^(2+cos n )

Relation between them ?

1 Answer

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1) This problem can be divided into 3 cases based on the range in which n lies and how trigonometric function varies within that range.3 cases are as follows:-

  1. ∀n,n∈[0, π/2 ], g(n)>f(n),and so f(n)=O(g(n))
  2. ∀n,n∈[π/2 , 3π/2 ],g(n)>f(n),and so f(n)=O(g(n))
  3. ∀n,n∈[π/2 , 3π/2 ],f(n)>g(n), and so f(n)= Ω(g(n))

And 1-3 will repeat for all subsequent values of sin and cos function as they are periodic in nature.So, from 1-3 we can observe that no specific relationship can be established between f(n) & g(n) as they behave differently in different ranges.

2)Same logic as 1

 

 

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