(Solved) : 2 Part Question Put Growth Rates Order Slowest Growing Fastest Answer F1 N F2 N Fk N Fi Q44018346 . . .
2. For each part of this question, put the growth rates inorder, from slowest-growing to fastest. That is, if your answer isf1(n), f2(n), . . . , fk(n), then fi(n) ≤ O(fi+1(n)) for all i. Iftwo adjacent ones are asymptotically the same (that is, fi(n) =Θ(fi+1(n))), you must specify this as well. Justify your answer(show your work). You may assume transitivity: if f(n) ≤ O(g(n))and g(n) ≤ O(h(n)), then f(n) ≤ O(h(n)), and similarly forlittle-oh, etc. (a) Polynomials. n + 1 n 4 1/n 1 n 2 + 2n − 4 n 2 √n 10100
(b) Logarithms and related functions. (log2 n) 2 log2 (n) log3(n) √ n log1.5 (n) log2 (n 2 )
(c) Logarithms in exponents. n log3 (n) n log2 n n 1/ log2 (n) n1
(d) Exponentials. Hint: Recall Stirling’s approximation, whichsays that n! ∼ n e n √ 2πn, i.e. limn→∞ n! ( n e ) n√ 2πn = 1. n!2n 2 2n 2 n log2 (n) 2 n+
2. For each part of this question, put the growth rates in order, from slowest-growing to fastest. That is, if your answer is fi (n), f2(n),…, fr(n), then f:(n) < 0(fi+1(n)) for all i. If two adjacent ones are asymptotically the same (that is, fi(n) = 0(fi+1(n))), you must specify this as well. Justify your answer (show your work). You may assume transitivity: if f(n) <0(g(n)) and g(n) < O(h(n)), then f(n) <O(h(n)), and similarly for little-oh, etc. (a) Polynomials. п? + 2n — 4 n2 1/n Vn 10100 n+1 (b) Logarithms and related functions. log, (n’) (log, n)² log, (n) log, (n) log15(n) (c) Logarithms in exponents. loga (n) log, n п n/ log2(n) п п (d) Exponentials. Hint: Recall Stirling’s approrimation, which says that n! ~ ()” /2an, i.e. limn+0 n! = 1. V2rn 22n 2n log2(n) 2n+7 2″ n! Show transcribed image text 2. For each part of this question, put the growth rates in order, from slowest-growing to fastest. That is, if your answer is fi (n), f2(n),…, fr(n), then f:(n)
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Answer to 2. For each part of this question, put the growth rates in order, from slowest-growing to fastest. That is, if your answ…
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