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 righthand 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 righthand side of the recurrence relation: 2(n 1)2 n n = (n 1)2 n + 2 n = n2 n, which is the lefthand 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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