(Solved):Let f(n)=0.5n^3, g(n) = 4n^2+2n. For what value of n is f(n) le g(n)? What does your answer have to do with the definitions of O and OMEGA? View Answer…

 

Question

Let {eq}f(n)=0.5n^3, g(n) = 4n^2+2n

{/eq}. For what value of n is {eq}f(n) le g(n)

{/eq}? What does your answer have to do with the definitions of O and OMEGA?

 

EXPERT ANSWER

Given:

{eq}f(n)=0.5n^3

g(n)=4n^2 + 2n

{/eq}

Now, using big-O notation.

{eq}f(n)leq g(n)\

text{then}, f(n)=O(g(n))

{/eq}

Find the value of “n”.

{eq}f(n)leq g(n)\

{/eq}

Using big-O

{eq}f(n)leq C.g(n)

{/eq}

{eq}5n^3leq C(4n^2+2n)

{/eq}

So, at C=1 and n=1 is the only value where equation satisfies.

Therefore, the value of n=1

{eq}f(n)leq g(n)

{/eq}

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