Discrete mathematics Week 5

Consider the nonhomogeneous linear recurrence relation an = 2an − 1 + 2n. Click and drag the given steps to their corresponding step names to show that an = n2n is a solution of the given recurrence relation in the correct order.

Step 1: If an=n2^n for all n, then an-1=(n-1)2^(n-1) Step 2:The right-hand side of the given recurrence relation can be written as 2(n-1)2^(n-1) +2^n Step 3: Substituting both equations into the recurrence, we obtain (n-1)2^n +2^n =n2^n

The Lucas numbers satisfy the recurrence relation Ln = Ln − 1 + Ln − 2, and the initial conditions are L0 = 2 and L1 = 1. Click and drag statements to find an explicit formula for the Lucas numbers.

Step 1: The Recurrence relation ln=ln-1+ln-2 has characteristic equation r^2 -r-1=0 Its roots are r=1sqrt5/2 Step 2: ln=a1(1sqrt5/2)^n +a2(1sqrt5/2)^n for some constants a1 and a2 Step 3: l0=2 and l1=1 to a1=1 and a2=1. ln(1sqrt5/2)^n +(1sqrt5/2)^n

Let an be the number of ways to climb n stairs if a person climbing the stairs can take one stair or two stairs at a time. Identify the initial condition for the recurrence relation in the previous question.

a0 = 1 and a1 = 1

Let an be the number of bit strings of length n that do not contain three consecutive 0s. Identify the initial conditions for the recurrence relation in the previous question.

a0 = 1, a1 = 2, and a2 = 4

Consider the nonhomogeneous linear recurrence relation an = 2an − 1 + 2n. Identify the solution of the given recurrence relation with a0 = 2.

an = (n + 2)2n

Solve these recurrence relations together with the initial conditions given. Identify the solution of the recurrence relation an = 6an − 1 - 8an − 2 for n ≥ 2 together with the initial conditions a0 = 4 and a1 = 10.

an = 3 · 2n + 4n

Let an be the number of bit strings of length n that do not contain three consecutive 0s. Identify a recurrence relation for an.

an = an - 1 + an - 2 + an - 3 for n ≥ 3

Let an be the number of ways to climb n stairs if a person climbing the stairs can take one stair or two stairs at a time. Identify a recurrence relation for an.

an = an −1 + an − 2 for n ≥ 2

Consider the nonhomogeneous linear recurrence relation an = 2an − 1 + 2n. Identify the set of all solutions of the given recurrence relation using the theorem given below. If {a(p)n} is a particular solution of the nonhomogeneous linear recurrence relation with constant coefficients an = c1an − 1 + c2an − 2 +· · ·+ckan − k + F(n), then every solution of the form {a(p)n + a(h)n}, where {a(h)n} is a solution of the associated homogeneous recurrence relation an = c1an − 1 + c2an − 2 +· · ·+ckan − k.

an = α(2)n + n(2)n

Let an be the number of ways to climb n stairs if a person climbing the stairs can take one stair or two stairs at a time. Identify the number of ways the person who can take one stair or two stairs at a time can climb a flight of eight stairs.

34

There are 345 students at a college who have taken a course in calculus, 214 who have taken a course in discrete mathematics, and 190 who have taken courses in both calculus and discrete mathematics. How many students have taken a course in either calculus or discrete mathematics?

369 Let C be the set of students who have taken a course in calculus; thus, |C| = 345. Let D be the set of students who have taken a course in discrete mathematics; thus, |D| = 214. Then, C ∩ Drepresents the set of students who have taken courses in both calculus and discrete mathematics; thus, |C ∩ D| = 190. Using the principle of inclusion-exclusion, we have |C ∪ D| = |C| + |D| - |C ∩ D|. By substituting the values of |C|, |D|, and |C ∩ D|, we get |C ∪ D| = 345 + 214 - 190 = 369

There are 347 students at a college who have taken a course in calculus, 214 who have taken a course in discrete mathematics, and 190 who have taken courses in both calculus and discrete mathematics. How many students have taken a course in either calculus or discrete mathematics?

371 |C ∪ D| = 347 + 214 - 190 = 371

Find the number of ways to distribute six different toys to three different children such that each child gets at least one toy. The number of ways to distribute six different toys to three different children such that each child gets at least one toy is

540 Evaluate the number of onto functions from a set with 6 elements (the toys) to a set with 3 elements (the children), since each toy is assigned to a unique child. There are 36 functions from a set with 6 elements (the toys) to a set with 3 elements (the children). There are C(3, 1)(3 − 1)6 functions from a set with 6 toys to a set with 3 children such that each child gets exactly one toy, C(3, 2)(3 − 2)6 functions from a set with 6 toys to a set with 3 children such that each child gets exactly two toys, and so on, with C(3, 3 − 1) · 16 functions from a set with 6 toys to a set with 3 children such that each child gets exactly 3 − 1 toys. Hence, by the inclusion-exclusion principle, there are 36 − C(3, 1)(3 − 1)6 + C(3, 2)(3 − 2)6 = 540 number of ways to distribute six different toys to three different children such that each child gets at least one toy.

Find the number of positive integers not exceeding 1000 that are not divisible by 3, 17, or 35.

610

Find the number of positive integers not exceeding 1010 that are not divisible by 3, 17, or 35.

616

Find the number of positive integers not exceeding 1020 that are not divisible by 3, 17, or 35.

621 We first compute the number of positive integers not exceeding 1020 that are divisible by at least one of 3, 17, and 35. We use the inclusion-exclusion principle on the three sets. Let A be the set of positive integers not exceeding 1020 that are divisible by 3, B be the set of positive integers not exceeding 1020 that are divisible by 17, and C be the set of positive integers not exceeding 1020 that are divisible by 35. We determine |A ∪ B ∪ C| as follows: |A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C| ∣∣A∪B∪C∣∣=⌊10203⌋+⌊102017⌋+⌊102035⌋−⌊10203·17⌋−⌊10203·35⌋−⌊102017·35⌋+⌊10203·17·35⌋|A∪B∪C|=10203+102017+102035-10203·17-10203·35-102017·35+10203·17·35 |A ∪ B ∪ C| = 340 + 60 + 29 - 20 - 9 - 1 + 0 = 399 The number of positive integers not exceeding 1020 that are divisible by at least one of 3, 17, and 35 is 399; so, the number of positive integers not exceeding 1020 that are not divisible by 3, 17, or 35 numbers is 1020 - 399 = 621.

Find the number of positive integers not exceeding 100 that are not divisible by 5 or by 7.

68

Find the number of positive integers not exceeding 104 that are not divisible by 5 or by 7.

72

Find the number of positive integers not exceeding 112 that are not divisible by 5 or by 7.

77 To find the number of positive integers not exceeding 112 that are not divisible by 5 or by 7, we will subtract from 112 the number of positive integers that are divisible. Of the positive integers not exceeding 112, there are ⌊112/5⌋1125 integers that are divisible by 5 and ⌊112/7⌋1127 that are divisible by 7. As 5 and 7 are relatively prime, the integers that are divisible by both 5 and 7 are those that are divisible by 5 · 7. Consequently, there are ⌊112/5·7⌋112/5·7 positive integers not exceeding 112 that are divisible by both 5 and 7. It follows that there are 112−⌊112/5⌋−⌊112/7⌋+⌊112/5·7⌋112-112/5-112/7+112/5·7 = 112 - 22 - 16 + 3 = 112 - 35 = 77 positive integers that are not divisible by 5 or by 7.

Let an be the number of bit strings of length n that do not contain three consecutive 0s. Identify the number of bit strings of length seven that do not contain three consecutive 0s.

81

Find the number of positive integers not exceeding 150 that are either odd or the square of an integer.

81 Clearly, there are 75 odd positive integers not exceeding 150 (half of the 150 numbers are odd), and there are 12 squares. Furthermore, half of these squares are odd. Thus, we compute the cardinality of the set in the question to be 75 + 12 - 6 = 81.

Solve these recurrence relations together with the initial conditions given. Arrange the steps to their corresponding step numbers to solve the recurrence relation an + 2 = -4an + 1+ 5an for n ≥ 0 together with the initial conditions a0 = 2 and a1 = 8.

Step 1: The characteristic equation and its roots are r2 + 4r - 5 = 0 and r = -5, 1, respectively. Step 2: The general solution is an = α1(-5)n + α21n = α1(-5)n + α2. Step 3: Using initial conditions, 2 = α1 + α2 and 8 = -5α1 + α2 Step 4: After solving, α1 = -1 and α2 = 3. Therefore, an = -(-5)n + 3.

Solve these recurrence relations together with the initial conditions given. Arrange the steps to solve the recurrence relation an = an − 2 for n ≥ 2 together with the initial conditions a0 = 5 and a1 = -1 in the correct order.

Step 1: r2 − 1 = 0; r = -1, 1 Step 2: an = α1(-1)n + α21n = α1(-1)n + α2 Step 3: 5 = α1 + α2 -1 = -α1 + α2 Step 4: α1 = 3 and α2 = 2Therefore, an = 3 · (-1)n + 2.

Solve these recurrence relations together with the initial conditions given. Arrange the steps to solve the recurrence relation an = an − 1 + 6an − 2 for n ≥ 2 together with the initial conditions a0 = 3 and a1 = 6 in the correct order.

Step 1: r2 − r − 6 = 0 and r = −2, 3 Step 2: an = α1(−2)n + α23n Step 3: 3 = α1 + α26 = −2α1 + 3α2 Step 4: α1 = 3 / 5 and α2 = 12 / 5Therefore, an = (3 / 5)(−2)n + (12 / 5)3n.

The Lucas numbers satisfy the recurrence relation Ln = Ln − 1 + Ln − 2, and the initial conditions are L0 = 2 and L1 = 1. Click and drag expressions to show that Ln = fn − 1 + fn + 1 for n = 2, 3, . . . , where fn is the nth Fibonacci number.

Step 1:Let P(n) be the proposition that ln=fn-1+fn+1 Step 2: Basic Step:P(1) is true because l2-l1 Step 3: Inductive Step: Assume P(j) is true for all j Step 4: Lk+Lk-1 STep 5: (fk-2+fk+1)+(fk+fk+1) Step 6: fk+fk+2