Use the definition of bigTheta to prove that 2n4 n3 5n2 4

Use the definition of big-Theta to prove that 2n^4 - n^3 - 5n^2 + 4/n^2 - 6n + 7 = Theta (n^2).

Solution

Answer:

f(n) = 2n^4 - n^3 - 5n^2 + 4 / n^2 -6n +7

g(n) = n^2

Now c1 *g(n) < = f(n) < = c2*g(n)

c1 * n^2 < = 2n^4 -n^3 -5n^2 +4 / n^2 -6n +7 < = c2 *g(n)

let c1 = 1 , c2 = 2

n^2 < = 2n^4 -n^3 -5n^2 +4 / n^2 -6n +7 < = 2n^2

now

2n^4 -n^3 -5n^2 +4 / n^2 -6n +7 < = 2n^2

2n^4 -n^3 -5n^2 + 4 = 2n^2(n^2 -6n +7)

2n^4 -n^3 -5n^2 +4 = 4n^4 -6n^3 +7n

2n^4 -4n^4 -n^3 +6n^3 - 5n^2 -7n + 4 = 0

=> -2n^4 + 5n^3 -5n^2 -7n +4 = 0

Here the average case of the term with lowest power of n = 2 therefore theta(n^2).