Discrete Mathematics: Homework 7 solution. Due:

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1 EE 2060 Discrete Mathematics spring 2011 Discrete Mathematics: Homework 7 solution Due: Let a n = 2 n n for n = 0, 1, 2,... (a) (2%) Find a 0, a 1, a 2, a 3 and a 4. (b) (2%) Show that a 2 = 5a 1 6a 0, a 3 = 5a 2 6a 1, and a 4 = 5a 3 6a 2. (c) (3%) Show that a n = 5a n 1 6a n 2 for all integers n with n 2. (a) We simply plug in n = 0, n = 1, n = 2, n = 3, and n = 4. Thus we have a 0 = = 6, a 1 = = 17, a 2 = = 49, a 3 = = 143, and a 4 = = 421. (b) Using our data from part (a), we see that 49 = , 143 = , and 421 = (c) This is algebra. The messiest part is factoring out a large power of 2 and a large power of 3. If we substitute n 1 for n in the definition we have a n 1 = 2 n n 1 ; similarly a n 2 = 2 n n 2. We start with the right-hand side of our desired identity: 5a n 1 6a n 2 = 5(2 n n 1 ) 6(2 n n 2 ) =2 n 2 (10 6) + 3 n 2 (75 30) =2 n n =2 n + 3 n 5 = a n 2. (a) (6%) Find a recurrence relation for the number of bit strings of length n that contain three consecutive 0s. (b) (1%) What are the initial conditions?

2 (c) (2%) How many bit strings of length seven contain three consecutive 0s? (a) Let a n be the number of bit strings of length n containing three consecutive 0 s. In order to construct a bit string of length n containing three consecutive 0 s we could start with 1 and follow with a string of length n 1 containing three consecutive 0 s, or we could start with a 01 and follow with a string of length n 2 containing three consecutive 0 s, or we could start with a 001 and follow with a string of length n 3 containing three consecutive 0 s, or we could start with a 000 and follow with any string of length n 3. These four cases are mutually exclusive and exhaust the possibilities for how the string might start. From this analysis we can immediately write down the recurrence relation, valid for all n 3 : a n = a n 1 + a n 2 + a n n 3. (b) There are no bit strings of length 0, 1, or 2 containing three consecutive 0 s, so the initial conditions are a 0 = a 1 = a 2 = 0 (c) We will compute a 3 through a 7 using the recurrence relation: a 3 = a 2 + a 1 + a = = 1 a 4 = a 3 + a 2 + a = = 3 a 5 = a 4 + a 3 + a = = 8 a 6 = a 5 + a 4 + a = = 20 a 7 = a 6 + a 5 + a = = 47 Thus there are 47 bit strings of length 7 containing three consecutive 0 s. 3. (a) (6%) Find a recurrence relation for the number of ways to climb n stairs if the person climbing the stairs can take one, two, or three stairs at a time. (b) (1%) What are the initial conditions? (c) (2%) How many ways can this person climb a flight of eight stairs?

3 (a) Let a n be the number of ways to climb n stairs. In order to climb n stairs, a person must either start with a step of one stair and then climb n 1 stairs or else start with a step of two stairs and then climb n 2 stairs or else start with a step of three stairs and then climb n 3 stairs. From this analysis we can immediately write down the recurrence relation, valid for all n 3 : a n = a n 1 + a n 2 + a n 3. (b) The initial conditions are a 0 = 1, a 1 = 1, a 2 = 2, since there is one way to climb no stairs, clearly only one way to climb one stair, and two ways to climb two stairs. (c) Each term in our sequence {a n } is the sum of the previous three terms, so the sequence begins a 0 = 1, a 1 = 1, a 2 = 2, a 3 = 4, a 4 = 7, a 5 = 13, a 6 = 24, a 7 = 44, a 8 = 81. Thus a person can climb a flight of 8 stairs in 81 ways under the restrictions in this problem. 4. A string that contains only 0s, 1s, and 2s is called a ternary string. (a) (6%) Find a recurrence relation for the number of ternary strings that contain two consecutive 0 s. (b) (1%) What are the initial conditions? (c) (2%) How many ternary strings of length six contain two consecutive 0 s? (a) Let a n be the number of ternary strings that contain two consecutive 0 s. To construct such a string we could start with either a 1 or a 2 and follow with a string containing two consecutive 0 s, or we could start with 01 or 02 and follow with a string containing two consecutive 0 s, we could start with 00 and follow with any ternary string of length n 2. Therefore the recurrence relation, valid for all n 2, is a n = 2a n 1 + 2a n n 2. (b) a 0 = a 1 = 0 (c) We will compute a 2 through a 6 using the recurrence relation: a 2 = 2a 1 + 2a = 1

4 a 3 = 2a 2 + 2a = 5 a 4 = 2a 3 + 2a = 21 a 5 = 2a 4 + 2a = 79 a 6 = 2a 5 + 2a = 281 Thus there are 281 bit strings of length 6 containing two consecutive 0 s. 5. Solve these recurrence relations together with the initial conditions given (a) (3%) a n = 7a n 1 10a n 2 for n 2, a 0 = 2, a 1 = 1. (b) (3%) a n = 2a n 1 a n 2 for n 2, a 0 = 4, a 1 = 1. (c) (3%) a n = 6a n 1 9a n 2 for n 2, a 0 = 3, a 1 = 3. (a) r 2 7r + 10 = 0 r = 2, 5 a n = α 1 2 n + α 2 5 n 2 = α 1 + α 2 1 = 2α 1 + 5α 2 α 1 = 3, α 2 = 1 a n = 3 2 n 5 n (b) r 2 2r + 1 = 0 r = 1, 1 a n = α 1 1 n + α 2 n1 n 4 = α 1 1 = α 1 + α 2 α 1 = 4, α 2 = 3 a n = 4 3n (c) r 2 + 6r + 9 = 0 r = 3, 3 a n = α 1 ( 3) n + α 2 n( 3) n 3 = α 1 3 = 3α 1 3α 2 α 1 = 3, α 2 = 2 a n = (3 2n)( 3) n

5 6. (4%) Find the solution to a n = 5a n 2 4a n 4 with a 0 = 3, a 1 = 2, a 2 = 6 and a 3 = 8. The characteristic equation is r 4 5r = 0. This factor as (r 2 1)(r 2 4) = (r + 1)(r 1)(r 2)(r + 2) = 0, so the roots are 1, 1, 2, 2. Therefore the general solution is a n = α 1 + α 2 ( 1) n + α 3 2 n + α 4 ( 2) n. Plugging in initial conditions gives 3 = α 1 + α 2 + α 3 + α 4, 2 = α 1 α 2 + 2α 3 2α 4, 6 = α 1 + α 2 + 4α 3 + 4α 4, and 8 = α 1 α 2 +8α 3 8α 4. The solution to this system of equations is α 1 = α 2 = α 3 = 1 and α 4 = 0. Therefore the answer is a n = 1 + ( 1) n + 2 n. 7. (5%) Solve the recurrence relation a n = 6a n 1 12a n 2 +8a n 3 with a 0 = 5, a 1 = 4, and a 2 = 88. The characteristic equation is r 3 6r r 8 = 0. By the rational root test, the possible rational roots are ±1, ±2, ±4. We find that r = 2 is a root. Dividing r 2 into r 3 6r 2 +12r 8, we find that r 3 6r 2 +12r 8 = (r 2) 3. Hence the only root is 2, with multiplicity 3, so the general solution is (by Thm.4) a n = α 1 2 n +α 2 n2 n +α 3 n 2 2 n. Plugging the initial conditions: 5 = a 0 = α 1 4 = a 1 = 2α 1 + 2α 2 + 2α 3 88 = a 2 = 4α 1 + 8α α 3 Solving this system of equations, we have α 1 = 5, α 2 = 1/2, and α 3 = 13/2. Therefore the answer is a n = 5 2 n + (n/2) 2 n + (13n 2 /2) 2 n = 5 2 n + n 2 n n 2 2 n (2%) What is the general form of the solutions of a linear homogeneous recurrence relation if its characteristic equation has the roots 1, 1, 1, 2, 2, 5, 5, 7. We can write down the general solution using Theorem 4 of Sec.7.2. In this case there are four distinct roots, so t = 4. The multiplicities are 3, 2, 2, and 1. So the general solution is a n = (α 1,0 +α 1,1 n+α 1,2 n 2 )( 1) n +(α 2,0 +α 2,1 n)2 n +(α 3,0 +α 3,1 n)5 n +α 4,0 7 n. 9. Consider the nonhomogeneous linear recurrence relation a n = 2a n n (a) (2%) Show that a n = n2 n is a solution of this recurrence relation.

6 (b) (2%) Use Theorem 5 of Sec.7.2 to find all solutions of this recurrence relation. (c) (2%) Find the solution with a 0 = 2. (a) We compute the right-hand side of the recurrence relation: 2(n 1)2 n n = (n 1)2 n + 2 n = n2 n, which is the left-hand side. (b) The solution of the associated homogeneous equation a n = 2a n 1 is easily found to be a n = α2 n. Therefore the general solution of the inhomogeneous equation is a n = α2 n + n2 n. (c) Plugging in a 0 = 2, we obtain α = 2. Therefore the solution is a n = 2 2 n +n2 n = (n + 2)2 n 10. What is the general form of the particular solution guaranteed to exist by Theorem 6 of the linear nonhomogeneous recurrence relation a n = 6a n 1 12a n 2 +8a n 3 +F (n) if (a) (2%) F (n) = 2 n (b) (2%) F (n) = ( 2) n (c) (2%) F (n) = n 3 ( 2) n (a) Since 2 is a root with multiplicity 3 of the characteristic polynomial of the associated homogeneous recurrence relation, Theorem 6 of Sec.7.2 tells us that the particular solution will be of the form n 3 p 0 2 n. (b) Since 2 is not a root of the characteristic polynomial of the associated homogeneous recurrence relation, Theorem 6 of Sec.7.2 tells us that the particular solution will be of the form p 0 ( 2) n. (c) Since 2 is not a root of the characteristic polynomial of the associated homogeneous recurrence relation, Theorem 6 of Sec.7.2 tells us that the particular solution will be of the form (p 3 n 3 + p 2 n 2 + p 1 n + p 0 )( 2) n. 11. Find a closed form for the generating function for the sequence {a n }, where

7 (a) (2%) a n = 2 n for n = 1, 2, 3, 4,... and a 0 = 0. (b) (2%) a n = 1/(n + 1)! for n = 0, 1, 2,... ( ) 10 (c) (2%) a n = for n = 0, 1, 2,... n + 1 (a) By Table 1 of Sec.7.4, the generating function for the sequence in which a n = 2 n for all n is 1/(1 2x). Here we can either think of subtracting out the missing constant term or factoring out 2x. Therefore the answer can be written as either 1(1 2x) 1 or 2x/(1 2x), which are of course algebraically equivalent. (b) The power series for the function e x is xn /n!. That is almost what we have here; the difference is that the denominator is (n + 1)! instead of n!. So (c) we have x n (n + 1)! = 1 x x n+1 (n + 1)! = 1 x by change of variable. This last sum is (e x 1), so our answer is (e x 1)/x. C(10, n + 1)x n = C(10, n)x n 1 = 1 x n=1 n=1 x n n! C(10, n)x n = 1 x ((1 + x)10 1) 12. For each of these generating functions, provide a closed formula for the sequence it determines. (a) (2%) (3x 1) 3 (b) (2%) x 2 (1 x) 3 (c) (2%) (1 + x 3 )/(1 + x) 3 (d) (2%) e 3x2 1 (a) First we need to factor out 1 and write this as (1 3x) 3. n=1 Then by the Binomial Theorem we get a n = C(3, n)( 3) n for n = 0, 1, 2, 3, and the

8 other coefficients are all 0. Alternatively, we could just multiply out this finite polynomial and note the nonzero coefficients: a 0 = 1, a 1 = 9, a 2 = 27, a 3 = 27. (b) We know that x 2 (1 x) 3 = x 2 (1 3x + 3x 2 x 3 ) = x 2 3x 3 + 3x 4 x 5, so we have a 2 = 1, a 3 = 3, a 4 = 3, a 5 = 1 (c) We split this into two parts : 1 (1 + x) + x 3 3 (1 + x) = ( 1) n C(n + 2, 2)x n + x 3 3 = = ( 1) n C(n + 2, 2)x n + ( 1) n C(n + 2, 2)x n + ( 1) n C(n + 2, 2)x n ( 1) n C(n + 2, 2)x n+3 ( 1) n 3 C(n 1, 2)x n Note that n and n 3 have opposite parities. Therefore a n = ( 1) n C(n + 2, 2) + ( 1) n 3 C(n 1, 2) = ( 1) n (C(n + 2, 2) C(n 1, 2)) = ( 1) n 3n for n 3 and a n = ( 1) n C(n + 2, 2) = ( 1) n (n + 2)(n + 1)/2 for n < 3. (d) e x = 1 + x + x 2 /2! + x 3 /3! +... It follows that n=3 e 3x2 = 1 + 3x 2 + (3x2 ) 2 2! + (3x2 ) 3 3! +... We can therefore read off the coefficients of the generating function for e 3x2 1. First, clearly a 0 = 0. Second, a n = 0 when n is odd. Finally, when n is even, we have a 2m = 3 m /m!. 13. Find the coefficient of x 12 in the power series of each of these functions. (a) (2%) 1/(1 2x) 2 (b) (2%) 1/(1 4x) 3 (a) The coefficient of x n in this power series is 2 n C(n + 1, 1). Thus the answer is 2 12 C(12 + 1, 1) = 53, 248.

9 (b) The coefficient of x n in this power series is 4 n C(n + 2, 2). Thus the answer is 4 12 C(12 + 2, 2) = 1, 526, 726, (5%) Use generating functions to find the number of ways to choose a dozen bagels from three varieties- egg, salty, and plain- if at least two bagels of each kind but no more than three salty bagels are chosen. The factors in the generating function for choosing the egg and plain bagels are both x 2 + x 3 + x The factor for choosing the salty bagels is x 2 + x 3. Therefore the generating function for this problem is (x 2 + x 3 + x 4 +..) 2 (x 2 + x 3 ). We want to find the coefficient of x 12, since we want 12 bagels. This is equivalent to finding the coefficient of x 6 in (1 + x + x ) 2 (1 + x). This function is (1 + x)(1 x) 2, so we want the coefficient of x 6 in 1/(1 x) 2, which is 7, plus the coefficient of x 5 in 1/(1 x) 2, which is 6. Thus the answer is (a) (4%) What is the generating function for {a k }, where a k is the number of solutions of x 1 + x 2 + x 3 + x 4 = k when x 1, x 2, x 3 and x 4 are integers with x 1 3, 1 x 2 5, 0 x 3 4, and x 4 1? (b) (2%) Use your answer to part(a) to find a 7 (a) The restriction on x 1 gives us the factor x 3 + x 4 + x The restriction on x 2 gives us the factor x + x 2 + x 3 + x 4 + x 5. The restriction on x 3 gives us the factor 1 + x + x 2 + x 3 + x 4. And the restriction on x 4 gives us the factor x + x 2 + x Thus the answer is the product of these: (x 3 + x 4 + x )(x + x 2 + x 3 + x 4 + x 5 )(1 + x + x 2 + x 3 + x 4 )(x + x 2 + x ) We can use algebra to rewrite this in closed form as x 5 (1 + x + x 2 + x 3 + x 4 ) 2 /(1 x) 2 (b) We want the coefficient of x 7 in this series, which is the same as the coefficient of x 2 in the series for (1 + x + x 2 + x 3 + x 4 ) 2 = 1 + 2x + 3x2 + higher order terms (1 x) 2 (1 x) 2

10 Since the coefficient of x n in 1/(1 x) 2 is n+1, our answer is = (5%) Use generating functions to solve the recurrence relation a k = 3a k k 1 with the initial condition a 0 = 1. Let G(x) = k=0 a kx k. Then xg(x) = k=0 a kx k+1 = k=1 a k 1x k. Thus G(x) 3xG(x) = a k x k k=0 k=1 3a k 1 x k = a 0 + k=1 k=0 (a k 3a k 1 )x k = 1 + k=1 4 k 1 x k = 1 + x 4 k 1 x k 1 = 1 + x 4 k x k 1 = 1 + x 1 4x = 1 3x 1 4x Thus G(x)(1 3x) = (1 3x)/(1 4x), so G(x) = 1/(1 4x). Therefore a k = 4 k. k=1

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