**P:1) **Show that f(n) +g(n) =O(n²) where f(n) =3n²-n+4 and g(n) =nlogn+5

__Solution__*:*

We know that, 'O' is called 'Big-oh' notation where O is defined as

f(n) =O(g(n)) only if there is positive constant C and where f(n) ≤c.g(n), n≥n_{0}

In the given problem, we have to prove that ,

f(n) +g(n) =O(n²)

where, f(n) =3n²-n+4 , g(n) =nlogn+5

Hence, we can write it as :

⇒3n²-n+4+nlogn+5 <4n² , where n≥5

Therefore, by taking n=5

⇒3(5)^{2} -5+4+5log5+5≤4(5)^{2}

⇒75-5+4+5log5+5≤100

⇒74+8.49≤100

⇒82.49≤100

f(n) +g(n) =O(n²)

Hence , proved.

**P:2) **Define the function f(n) =12n^{2}+6n is O(n^{3}) and y(n) .

__Solution__:

Given that,

f(n) =12n^{2}+6n

Let, k>0 be a constant, Consider n_{0}=(12+6) /k

⇒n≥n_{0}

⇒kn^{3}≥12n^{3}+6n^{2}≥12n^{2}+6n

therefore, f(n)=O(n³)

To show, that f(n) =y(n) let k>0 be any constant.

Consider, n_{0}=k/12 , for all n≥n_{0}

⇒12n^{2}+6n≥12n^{2}≥kn

Therefore, f(n) =y(n)

**P:3) **Show that f(n)=4n^{2}-64n+288=y(n^{2})

__Solution__:

Given that:

f(n)=4n^{2}-64n+288

We need to prove

4n^{2}-64n+288=y(n^{2})

we can prove that by writing above equation in following form,

4n^{2}-64n+288≥n^{2}

where, n≥1(As per the definition of y)

⇒4(1)^{2}-64(1)+288≥(1)^{2}

^{⇒4-64+288≥1}

^{⇒228≥1}

^{Therefore, 4n²-64n+288=y(n²)}

^{Hence,proved.}

**P:4)** Find big-oh notation and little oh notation for ,f(n)=7n^{3}+50n^{2}+200

__Solution__:

Given that ,

f(n)=7n^{3}+50n^{2}+200

According to Big-oh notation

f(n)≤k.g(n)

|7n^{3}+50n^{2}+200|≤7n^{3}+50n^{2}+200

⇒|7n^{3}+50n^{2}+200|≤7n^{3}+50n^{3}+200n^{3}

⇒|7n^{3}+50n^{2}+200|≤257n^{3}

⇒|7n^{3}+50n^{2}+200|≤257|n^{3}|

⇒f(n)=7n^{3}+50n^{2}+200

⇒k=257

⇒g(n)=n^{3}

Therefore, 7n^{3}+50n^{2}+200=O(n^{3})

The liitle oh notation is f(n)=O(n^{4})