Recurrence Assuming in each case that T(n) is eventually nondecreasing, use Theo
ID: 3938144 • Letter: R
Question
Recurrence
Assuming in each case that T(n) is eventually nondecreasing, use Theorem B.5 to determine the order of the following recurrence equations. T(n) = 2T(n/5) + 6n^3 for n > 1, n a power of 5 T(1) = 6 T(n) = 40T(n/3) + 2n^3 for n > 1, n a power of 3 T(1) = 5 Suppose a complexity function T(n) is eventually nondecreasing and satisfies where b greaterthanorequalto 2 and k greaterthanorequalto 0 are constant integers, and a, c, and d are constants such that a > 0, c > 0, and d greaterthanorequalto 0. T(n) belongs to {theta(n^k) if a b^k. Furthermore, if, in the statement of the recurrence, T(n) = aT(n/b) + cn^k is replaced by T(n) lessthanorequalto aT(n/b) + cn^k or T(n) greaterthanorequalto aT(n/b) + cn^k, then Result B.5 holds with "big O" or ohm, respectively, replacing theta.Explanation / Answer
a)Comparing (a) with Master theorem,a=2,b=5,c=6,k=3
2< 53
i.e. a<b3
Hence ,Order is O(n3)
b)
Comparing (b) with Master theorem,a=40,b=3,c=2,k=3
40>33
i.e.a>bk
O(nlog3(40))
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