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Question
Submitted by azi.vb on Sun, 2012-05-20 18:29
due on Sun, 2012-05-20 18:25
answered 1 time(s)
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azi.vb bought 8 out of 10 answered question(s)

question about a big oh notation

Q1) Show, by applying the definition of the O-notation, that each of the

following is true.

- If f(n)= n(n-1)/2, then f(n) = O(n^2).

- If f(n)= n+ log n, then f(n) = O(n).

- 1+ n+ n^2 + n^3 = O(n^3).

Q2) State without proof whether each of the following is True or False.

- 7 = O(1).

- n + n^4 = O(n^3).

- For any polynomial T(n), T(2n) = O(T(n)).

- For any function T(n), T(2n) = O(T(n)).

Q3) Show, by the definition of the O-notation, that n^3 != O(n^2).

(Note != means not-equal.)

Q4) Let T1(n)= O(f(n)) and T2(n)= O((g(n)). Prove by the definition of

the O-notation, this implies T1(n) + T2(n)= O(f(n) + g(n)).

Q5) Let T1(n)= O(f(n)) and T2(n)= O((g(n)). Prove by the definition of

the O-notation, this implies T1(n) * T2(n)= O(f(n) * g(n)).

Answer
Submitted by zext on Sun, 2012-05-20 22:37
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Q1) xxxxx by xxxxxxxx xxx definition xx the xxxxxxxxxxx that each xx the following xx xxxxx

x xx f(n)= xxxxxxxxx xxxx f(n) x O(n^2). - xx f(n)= n+ log xx xxxx xxxx = xxxxxxx 1+ n+ xxx x xxx = O(n^3).

xxxxxxxx

F(n) = n(n-1)/2 < xxx - xx = n^2 – n < n^2 xxx all x >x

Hence, f(n) < n^2 xxx xxx n > 1

xxxxxx by definition of xxxxxxxxxx , f(n) x xxxxxx

xxxx x n x logn xxx xx xxxx xxxx x > xxxxxx xxx xxx n > x {this xx xxxxxxx e^n > n and by taking log xx xxxx xxxxx we xxxx x > logn}

xxxxxx f(n) < n x x = 2n == > xxxx < xx for xxx x>x where C = 2

Hence, by definition xx xxxxxxxxxx, f(n) = O(n)

xxxxxxxx (1 x x + n^2 x xxxx < xxxx + n^3 + n^3 x n^3) xxx all x>1

Hence, xx + x + n^2 + n^3) < xxxx for xxx n > x where x x 4

Hence, xx definition of xxxxxxxxxxx (1 + x x xxx

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