Asymptotic Notations

Question

Consider the following statements:
$(1)$ Any two functions $f,g$ are always comparable under big Oh,that is $f=O(g)$ or $g=O(f)$
$(2)$ If $f=O(g)$ and $f=O(h)$ then $g(n)=\theta(h)$
$A)$ $(1)$ is true $(2)$ is false
$B)$ $(1)$ is false $(2)$ is true
$C)$ Both are false
$D)$ Both are true

Comments

Both are false

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o sorry wrong example ........ example of somoshree is right

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Option (C) is the correct

See here

https://cs.stackexchange.com/questions/1780/are-the-functions-always-asymptotically-comparable

Answer 1

The first statement is false. 
Take two functions, $f(n) = 0.5$, $g(n) = sin(n)$.

You cannot compare these two functions using Big-Oh notation.

For more, read here: https://cs.stackexchange.com/questions/1780/are-the-functions-always-asymptotically-comparable

Answer 2

Statement 1 is false.

Consider f(n) = 0.5 and g(n) = sin(n)
or, consider f(n) = n and g(n) = 1 when n is even; $n^{2}$ when n is odd.

In both the above cases neither f(n) = Og(n) nor g(n) = Of(n)

 

Statement 2 is false

Consider g(n) = 2n and h(n) = n and f(n) = 10n

 

Other important statements

All the below cases are possible:-

$f(n) = Og(n)$ and $g(n) = Of(n)$ — Case 1

$f(n) = Og(n)$ and $g(n) \neq Of(n)$ — Case 2

$f(n) \neq Og(n)$ and $g(n) \neq Of(n)$ — Case 3

 

Case 1

When f(n) and g(n) are identical both functions can upper bound each other.

 

Case 2

f(n) = n and g(n) = $n^{2}$

 

Case 3

See "Statement 1 is false"