This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 6. Let us denote: a = (log(C1)log6)/ (log31), b = log31. Note that 1<b<2 since log4=2, log2=2. Then we have: n > a + (1/b) logn. The above is clearly true if n satisfies: n/2 > a and n/2 > (1/b) logn. The first inequality implies n > 2a, the second is equivalent to: (b/2)n > logn. In the question 1d above we have proved that logn 2 n 1/2 . We can now consider (b/2)n > n/2 > 2 n 1/2 . But n/2 > 2 n 1/2 is equivalent to n 1/2 > 4, i.e. n>8. Hence for n>8 we have (b/2)n > n/2 > 2 n 1/2 > logn. This means n > max(8,2a)=max(8,2(log(C1)log6)/ (log31)). c.[3] n(n1)/2 = O(n). We have to find such n (not necessarily the smallest one) that for every n n n(n1)/2 > Cn. If n > 2C+1, then n 2 > (2C+1)n = 2Cn+n n 2n > 2Cn n(n1)/2 = (n 2n)/2 > Cn d.[5] n+1 = O(log n). We have to find such n (not necessarily the smallest one) that for every n n...
View
Full
Document
This note was uploaded on 12/10/2009 for the course CAS 2c03 taught by Professor Janicki during the Spring '03 term at McMaster University.
 Spring '03
 janicki

Click to edit the document details