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# Note for Discrete Mathematics - DMS By vtu rangers

• Discrete Mathematics - DMS
• Note
• Visvesvaraya Technological University Regional Center - VTU
• 9 Topics
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12. MATHEMATICAL INDUCTIONDiscrete Mathematics Mathematical induction, is a technique for proving results or establishing statements for natural numbers. This part illustrates the method through a variety of examples. Definition Mathematical Induction is a mathematical technique which is used to prove a statement, a formula or a theorem is true for every natural number. The technique involves two steps to prove a statement, as stated below: Step 1(Base step): It proves that a statement is true for the initial value. Step 2(Inductive step): It proves that if the statement is true for the n number n), then it is also true for (n+1) th th iteration (or iteration ( or number n+1). How to Do It Step 1: Consider an initial value for which the statement is true. It is to be shown that the statement is true for n=initial value. Step 2: Assume the statement is true for any value of n=k. Then prove the statement is true for n=k+1. We actually break n=k+1 into two parts, one part is n=k (which is already proved) and try to prove the other part. Problem 1 n 3 -1 is a multiple of 2 for n=1, 2, ... Solution 1 Step 1: For n=1, 3 -1 = 3-1 = 2 which is a multiple of 2 n k Step 2: Let us assume 3 -1 is true for n=k, Hence, 3 -1 is true (It is an assumption) We have to prove that 3 3 k+1 k k+1 -1 is also a multiple of 2 k k – 1 = 3 × 3 – 1 = (2 × 3 ) + (3 –1) k k The first part (2×3 ) is certain to be a multiple of 2 and the second part (3 -1) is also true as our previous assumption. Hence, 3 k+1 – 1 is a multiple of 2. n So, it is proved that 3 – 1 is a multiple of 2. Problem 2 2 1 + 3 + 5 + ... + (2n-1) = n for n=1, 2, ... 48

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Discrete Mathematics Solution 2 Step 1: For n=1, 1 = 1 , Hence, step 1 is satisfied. Step 2: Let us assume the statement is true for n=k. 2 Hence, 1 + 3 + 5 + ... + (2k-1) = k is true (It is an assumption) 2 We have to prove that 1 + 3 + 5 + ... + (2(k+1)-1) = (k+1) also holds 1 + 3 + 5 + ... + (2(k+1) – 1) = 1 + 3 + 5 + ... + (2k+2 – 1) = 1 + 3 + 5 + ... + (2k + 1) = 1 + 3 + 5 + ... + (2k – 1) + (2k + 1) 2 = k + (2k + 1) = (k + 1) 2 2 So, 1 + 3 + 5 + ... + (2(k+1) – 1) = (k+1) hold which satisfies the step 2 2. Hence, 1 + 3 + 5 + ... + (2n – 1) = n is proved. Problem 3 n n n Prove that (ab) = a b is true for every natural number n Solution 1 1 1 Step 1: For n=1, (ab) = a b = ab, Hence, step 1 is satisfied. k k k Step 2: Let us assume the statement is true for n=k, Hence, (ab) = a b is true (It is an assumption). We have to prove that (ab) k k+1 =a k+1 k+1 b also hold k k Given, (ab) = a b Or, (ab) (ab)= (a b ) (ab) [Multiplying both side by ‘ab’] Or, (ab) k k+1 k k k k = (aa ) ( bb ) k+1 (ab) Or, = (ak+1bk+1) Hence, step 2 is proved. n n n So, (ab) = a b is true for every natural number n. Strong Induction Strong Induction is another form of mathematical induction. Through this induction technique, we can prove that a propositional function, P(n) is true for all positive integers, n, using the following steps:     Step 1(Base step): It proves that the initial proposition P(1) true. Step 2(Inductive step): It proves that the conditional statement   [ (1) ⋀ (2) ⋀ (3) ⋀ … … … … ⋀ ( )] → ( + 1) is true for positive integers k. 49

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13. RECURRENCE RELATIONDiscrete Mathematics In this chapter, we will discuss how recursive techniques can derive sequences and be used for solving counting problems. The procedure for finding the terms of a sequence in a recursive manner is called recurrence relation. We study the theory of linear recurrence relations and their solutions. Finally, we introduce generating functions for solving recurrence relations. Definition A recurrence relation is an equation that recursively defines a sequence where the next term is a function of the previous terms (Expressing F n as some combination of Fi with i<n). Example: Fibonacci series: Fn = Fn-1 + Fn-2, Tower of Hanoi: Fn = 2Fn-1 + 1 Linear Recurrence Relations A linear recurrence equation of degree k is a recurrence equation which is in the format xn= A1 xn-1+ A2 xn-1+ A3 xn-1+... Ak xn-k (An is a constant and Ak≠0) on a sequence of numbers as a first-degree polynomial. These are some examples of linear recurrence equations: Recurrence relations Fn = Fn-1 + Fn-2 Initial values Solutions a1=a2=1 Fibonacci number Fn = Fn-1 + Fn-2 a1=1, a2=3 Lucas number Fn = Fn-2 + Fn-3 a1=a2=a3=1 Padovan sequence Fn = 2Fn-1 + Fn-2 a1=0, a2=1 Pell number How to solve linear recurrence relation Suppose, a two ordered linear recurrence relation is: F n = AFn-1 +BFn-2 where A and B are real numbers. The characteristic equation for the above recurrence relation is: 2 x − Ax − B = 0 Three cases may occur while finding the roots: Case 1: If this equation factors as (x- x1)(x- x1) = 0 and it produces two distinct real n n roots x1 and x2, then Fn = ax1 + bx2 is the solution. [Here, a and b are constants] 2 Case 2: If this equation factors as (x- x1) = 0 and it produces single real root x1, then n n Fn = a x1 + bn x1 is the solution. Case 3: If the equation produces two distinct real roots x1 and x2 in polar form x1 = r ∠ θ and x2 = r ∠(- θ), then Fn = rn (a cos(nθ)+ b sin(nθ)) is the solution. 50

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Discrete Mathematics Problem 1 Solve the recurrence relation Fn = 5Fn-1 - 6Fn-2 where F0 = 1 and F1 = 4 Solution The characteristic equation of the recurrence relation is: 2 x – 5x + 6=0, So, (x-3) (x-2) = 0 Hence, the roots are: x1 = 3 and x2= 2 The roots are real and distinct. So, this is in the form of case 1 Hence, the solution is: n Fn = ax1 + bx2 n n n Here, Fn = a3 + b2 (As x1 = 3 and x2= 2) Therefore, 0 0 1 1 1=F0 = a3 + b2 = a+b 4=F1 = a3 + b2 = 3a+2b Solving these two equations, we get a = 2 and b = -1 Hence, the final solution is: n n n Fn = 2.3 + (-1) . 2 = 2.3 - 2 n Problem 2 Solve the recurrence relation Fn = 10Fn-1 - 25Fn-2 where F0 = 3 and F1 = 17 Solution The characteristic equation of the recurrence relation is: 2 x –10x -25 =0, So, 2 (x – 5) = 0 Hence, there is single real root x1 = 5 As there is single real valued root, this is in the form of case 2 Hence, the solution is: n Fn = ax1 + bnx1 0 n 0 3 = F0= a.5 + b.0.5 = a 1 1 17 = F1= a.5 + b.1.5 = 5a+5b Solving these two equations, we get a = 3 and b = 2/5 51