hw5solutions

hw5solutions - EE 369 Homework #5 Solutions (Sixth Edition)...

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Unformatted text preview: EE 369 Homework #5 Solutions (Sixth Edition) Section 2.4 10) T(1)=1 T(2)=2 T(3)=3 T(n)= T(n-1)+2T(n-2)+3T(n-3) n>3 Solution: 1, 2, 3, 10, 22 12) Prove the property of Fibonacci numbers: F(n)=5F(n-4)+3F(n-5) for n>=6 F(n) = F(n-2) + F(n-1) = F(n-3) + F(n-4) + F(n-2) + F(n-3) = 2F(n-3) + F(n-4) + F(n-2) = 2[F(n-4) + F(n-5)] + F(n-4) + [F(n-3) + F(n-4)] = 5F(n-4)+ 3F(n-5) 17) Prove the property of the Fibonacci numbers for n>=1 F(2)+F(4)+....+F(2n)=F(2n+1)-1 n=1: F(2)=F(3)-1 or 1=2-1 (true) Assume true for n=k: F(2)+ ... F(2k) = F(2k+1)-1 Show true for n=k+1: F(2)+ ... F(2(k+1)) = F(2(K+1)+1) -1 F(2)+....+F(2(k+1)) = F(2)+....F(2k)+F(2(k+1)) = F(2k+1)-1 + F(2(K+1)) inductive hypothesis = F(2k+3) -1 recurrence relation = F(2(K+1)+1)-1 46) Give a recursive definition for the set of all strings of well balanced parenthesis. 1. A string without parenthesis is well balanced. 2. If A and B are strings of well balanced parentheses, so are (A) and AB. Section 2.5 2. Solve the recurrence relation subject to the basis step: F(1)=2 F(n)=2F(n-1) + 2^n for n>=2 The recurrence relation matches Equation(6) with c=2 and g(n) = 2^n. From equation(8) the solution is F(n) = 2^(n-1)(2) + \sum(i=2 to n) 2^(n-i)2^i = 2^n + \sum(i=2 to n) 2^n = n(2^n) 3. T(1)=1 T(n)=2T(n-1) + 1 for n>=2 The recurrence relation matches Equation(6) with c=2 and g(n) = 1. From equation(8) the solution is T(n) = 2^(n-1)(1) + \sum(i=2 to n) 2^(n-i)(1) = 2^(n-1) + ...+2+1 = 2^n - 1 8. Solve the recurrence relation subject to the basis step by using the expand, guess, and verify approach. S(1)=1 S(n)=nS(n-1)+n! S(n)=nS(n-1)+n! =n[(n-1)S(n-2)+(n-1)!]+n!...
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This note was uploaded on 09/24/2009 for the course ECE 369 taught by Professor Bagchi during the Spring '08 term at Purdue University.

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hw5solutions - EE 369 Homework #5 Solutions (Sixth Edition)...

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