Question

Solve the following using iteration method. Note: T(1) = 1. 2. recurrences GE) T(п) 2T 2.1 3 Т(п) 2T (п — 2) + 5 2.2 Solve the following using Master Theorem. 3. recurrenсes T(п) log n n 4T .3 3.1 n 5T 2 n2 log n T(п) 3.2

Answer 1

2 1TC 2T( UA - 22-) 2T) 4 3 2 3 2(24) 23T(4) 2 3-4 2-3 22.3 2-2 2 = 2 (2-T) 3.43 22-3 23 3 2-3 2-3 3 23.3 k fimes t 2-3 K2 2 n TA will stop when 1 n d4 rt 2 2 2 decreauing bup Aways decreasing bP ocu Dn

2-2). Tln) 2T7N-2) + 5 =2(2TM-4)t 5)+ 5 22Tn-4) 227n-6)t 5)5.2 t 5 237-6522 K times 2TIM-2-K) 5.2 tS K-2 +5-2 + 5.2 5.2 5 Tt wil stop tohen n-2K- L k-2 2 52 (1-2 2 2- KLDtL 2 (T 25.2 2 5 K 6.22-5 D(2 2

3). Maiter-the orem- benea Re.cwrence elafions solvable by Madet's heorem is Tin) aTinb)in) 3 cases may arwve n9^a>fin) reatex j polynomiof funton atteact then Tin)- Bntoja) . fn) Catteast polynomiatty gneoter Ihen Tin) then Tin) eCfenogn) 3. Tin) W2) neog n + ЕВoye o9вода fin togn Crany 53 2o, Ten)= (ntogst ebgn 3.2 TnJ 5T2 n0gn 2-32 GTВои 7ъЧВазh finE nAgn fenJ nogha Zo, Tn OngL)B(n02|