Consider the following recurrence relation. if n = 0 if n > 0 Q(n) - {2. (2. Q(n-1)-3 Prove by induction that Q(n) = 2 (Induction on n.) Let f(n) = 2" + 3. Base Case: If n = 0, the recurrence relation says that Q(0) = 4, and the formula says that f(0) = + 3, for all n ≥ 0. Inductive Hypothesis: Suppose as inductive hypothesis that Q(k-1)= F(k-1) Inductive Step: Using the recurrence relation, Q(K) = 2 Q(k-1)-3, by the second part of the recurrence relation 2-1+3 = (2k + 6)-3 -2²+3 - 3, by inductive hypothesis so, by induction, Q(n) = f(n) for all n ≥ 0. X for some k> 0. , so they match.

Consider the following recurrence relation. if n = 0 if n > 0 Q(n) - {2. (2. Q(n-1)-3 Prove by induction that Q(n) = 2 (Induction on n.) Let f(n) = 2" + 3. Base Case: If n = 0, the recurrence relation says that Q(0) = 4, and the formula says that f(0) = + 3, for all n ≥ 0. Inductive Hypothesis: Suppose as inductive hypothesis that Q(k-1)= F(k-1) Inductive Step: Using the recurrence relation, Q(K) = 2 Q(k-1)-3, by the second part of the recurrence relation 2-1+3 = (2k + 6)-3 -2²+3 - 3, by inductive hypothesis so, by induction, Q(n) = f(n) for all n ≥ 0. X for some k> 0. , so they match.

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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