91Clearly this is smaller than√nlogn, so we will guess thatT(n) = Θ(√nlogn.To complete the proof we must show thatT(n)is inΩ(√nlogn), but thisturns out to be impossible.On the other hand, the recurrence is clearlyΩ√n. So, let’s guess thatT(n)is inO(√n)forc= 4, n0= 2. We prove this by induction.Base case:T(2) = 1 +√2<4√2, so the hypothesis is correct.Induction Hypothesis: For any value less than or equal tok,T(k)<4√k.Induction Step: For2k,T(2k) =T(k) +√2k.By the induction hypothesis,T(k) +√2k <4√k+√2k.For the theorem to be correct,T(k) +√2k <4√k+√2k <4√2kwhich is true. Thus, by the principle of Mathematical Induction, the theoremis correct.14.11Expanding the recurrence, we getT(n)=2T(n/2) +n
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