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Unformatted text preview: 6. Let us denote: a = (log(C-1)-log6)/ (log3-1), b = log3-1. 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(C-1)-log6)/ (log3-1)). c. n(n-1)/2 = O(n). We have to find such n (not necessarily the smallest one) that for every n n n(n-1)/2 > Cn. If n > 2C+1, then n 2 > (2C+1)n = 2Cn+n n 2-n > 2Cn n(n-1)/2 = (n 2-n)/2 > Cn d. n+1 = O(log n). We have to find such n (not necessarily the smallest one) that for every n n...
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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