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Prove that f ( n ) = 3 n 2 + 6 n + 9 f ( n ) = 3 n 2 + 6 n + 9 f( n ) = 3n^2+6n...

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Prove that f(n)=3n2+6n+9f(n)=3n2+6n+9f( n ) = 3n^2+6n+9 is not O(n)O(n)O( n )

Matching each statement with the correct sequence in a formal proof.

To prove f(n)f(n)f( n ) is not O(n)O ( n ), we use a proof by contradiction and we assume f(n)f( n ) is O(n)O ( n ).

We consider a specific integer mm such that m>k>0m>k>0 and 3m>C3m>C, then using the UI rule, we have f(m)=3m2+6m+9<Cmf( m ) = 3m^2+6m+9< C m.

The assumption is not valid so f(n)f( n ) is not O(n)O ( n )

proof

3m<C3m < C is contradicting to the statement that 3m>C3m > C.

Since m>k>0m>k>0, we can derive 3m2<3m2+6m+9<Cm3m^2 < 3m^2+6m+9 < Cm, so 3m<C3m < C.

QED

Using the EI rule, there exist positive integers CC and kk, such that ∀n((n>k)→(f(n)<Cn))\forall n ((n > k) \to (f( n ) < C n)), where n∈Zn \in \mathbb{Z}.

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