MAT 243 Written Work.pdf - 1 f(x = x5 x3 logx x2 1 5 5 x5...

This preview shows page 1 - 2 out of 2 pages.

1. f(x) = x +1 2 x + x logx 5 3 overtakes , so f(x) = = = x 5 x 5 x +1 2 x 2 x 5 x 3 Therefore, since n is equal to 3, O( ) = O( ) x n x 3 3 is the smallest integer n. 2. ln ( x ) x 2 Using the limit theorem: O [ ] = O [ ] * O [ ] ( x ) f ( x ) f 1 * 2 ( x ) f 1 f ( x ) 2 So if and , then = O( ) and x f 1 = 2 ln ( x ) f 2 = f 1 x 2 O ( x ) f 2 = Next, O( ln(x)) = O( ) * O(x) = O( ) x 2 x 2 x 3 Therefore is big-O of ln ( x ) x 2 x 3 3. By definition of O(n), we must prove that for some n ( ln n 0 n 1 n n 2 +1) n 3 n 2 2 . n 3
6 A) No, there is no smallest real number a for which is big-O of . 6 x 2 x a x B) Yes, there is a smallest integer such is big-O of . If we pick a=6, then | , 6 x 2 x a x | ≤ | x 6 | a x 2 x meaning that is 0( ). 6 x 2 x a x 7. O means there are positive constants

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture