MATH 242 – Principles of Analysis Solutions for Problem Set 2 – due: Feb. 8

‘A‘ Section

1. Let x ∈ [−1, 2]. Determine the largest and smallest values of |x − 5|, |x + 5|. Solution: If x ∈ [−1, 2], then x − 5 ∈ [−6, −3], so the largest and smallest values of |x − 5| are 6 and 3 respectively. (You can also see these as the distances from −1 to 5 and 2 to 5 along the number line, thinking geometrically.) Next, x + 5 ∈ [4, 7], so the largest and smallest values of |x + 5| are 7 and 4 respectively.

2. Use the binomial theorem (Theorem 1.4.1) for all parts of this problem. a. Expand using the binomial theorem and simplify as much as possible:

(a^{2} + 3b)^{5}.

Solution: We have

(a^{2} + 3b)^{5} = a^{10} + 15a^{8}b + 90a^{6}b^{2} + 135a^{4}b^{3} + 405a^{2}b^{4} + 243b^{5}.

b. What is the coefficient of x^{3} in the expansion of

(\frac{x^{5} + 3}{x^{2}})^{3} .

Solution: This coefficient is zero. The powers of x actually appearing in the binomial expansion will be (x^{5})^{3} = x^{15}, (x^{5})^{2}x^{−2} = x^{8}, x^{5}(x^{−2})^{2} = x and (x^{−2})^{3} = x^{−6}.

c. What is ∑_{k=0}^{n} \binom{n}{k}2^{k}? Explain. Solution: From the binomial theorem this sum is what we obtain from

(1 + 2)^{n} = 3^{n}.

d. What is ∑_{k=0}^{n}(−1)^{k}\binom{n}{k}? Explain.

Solution: From the binomial theorem this sum is what we obtain from

(1 − 1)^{n} = 0^{n} = 0.

3. For each of the following statements, say whether the statement is true or false. If it is false, give a counterexample; if it is true, give a short reason. a. A set A ⊂ R is bounded if there exists some B > 0 such that |x| ≤ B for all

x ∈ A. Solution: TRUE – the set is bounded above by B and below by −B.

b. If A, B ⊂ R are bounded, then A ∩ B is also bounded.

Solution: TRUE – If A is bounded above by M_{A} and B is bounded above by M_{B}, then A ∩ B is bounded above by min(M_{A}, M_{B}). Similarly, if A is bounded below by m_{A} and B is bounded below by m_{B}, then A ∪ B is bounded above by max(m_{A}, m_{B}).

c. If A, B ⊂ R are bounded, then D = {x + y | x ∈ A, y ∈ B} is also bounded. Solution: TRUE – Suppose A is bounded above by M_{A} and B is bounded above by M_{B}. Similarly, suppose A is bounded below by m_{A} and B is bounded below by m_{B}. Then for all x ∈ A and y ∈ B we have m_{A} ≤ x ≤ M_{A} and m_{B} ≤ y ≤ M_{B}. It follows that m_{A} + m_{B} ≤ x + y ≤ M_{A} + M_{B}. Therefore D is bounded.

d. If A, B ⊂ R_{>0} are bounded, then Q = {y/x | x ∈ A, y ∈ B} is also bounded. Solution: FALSE – Let A = (0, 1) and let B = {1}. The set Q is the set {1/y | y ∈ (0, 1)} which is not bounded above.

4.

a. Let A = [0, 3) ∩ (2, 5]. What is a = lub A? What is b = glb A? Are a, b ∈ A? Solution: We have A = (2, 3). Therefore a = 3, which is not in A. Similarly, b = 2 is not in A either.

b. Let B = {x ∈ R | 0 ≤ x^{2} − 2x + 1 ≤ 1}. What is a = lub B? What is b = glb B? Are a, b ∈ B? Solution: B is the set where (x − 1)^{2} ≤ 1, or x ∈ [0, 2], so a = 2 and b = 0 are both in B.

‘B‘ Section

1. Let x, y be any real numbers.

a. Show that |x| − |y| ≤ |x − y| and deduce that ||x| − |y|| ≤ |x − y|.

Solution: From the usual triangle inequality,

|x| = |(x − y) + y| ≤ |x − y| + |y|.

Subtracting, we obtain |x| − |y| ≤ |x − y| as desired. Similarly, reversing the roles of x, y, we have |y| − |x| ≤ |y − x| = |x − y|. Since either |x| ≥ |y| or |y| ≥ |x| is true, we have either ||x| − |y|| = |x| − |y| or ||x| − |y|| = |y| − |x|. Since both of those are ≤ |x − y|, it follows that ||x| − |y|| ≤ |x − y| as desired. b. Show that if x, y > 0, then x < y is equivalent to x^{n} < y^{n} for all n ≥ 1

Solution: ⇒: We argue by induction on n. The base case is the same as the hypothesis so there is nothing to prove. Assume x^{k} < y^{k} for some positive integer k. Then since x > 0, we can multiply by x on both sides to get x^{k+1} < y^{k}x. Similarly, we can multiply the base case x < y by y^{k} > 0 on both sides to get xy^{k} < y^{k+1}. But then transitivity of the order relation implies x^{k+1} < y^{k+1}. This shows that x^{n} < y^{n} for all n ≥ 1 by induction. Conversely, if x^{n} < y^{n} for all n ≥ 1, then x < y directly from the case n = 1 (!)

c. Show that if 0 < x < y, then \sqrt{y} − \sqrt{x} < \sqrt{y − x}. Solution: By part b with n = 2, since \sqrt{y} − \sqrt{x} > 0 and \sqrt{y − x} > 0, it suffices to show that

(\sqrt{y} − \sqrt{x})^{2} < (\sqrt{y − x})^{2}.

But the left side here is y − 2\sqrt{y}\sqrt{x} + x and the right side is y − x. We have

(y − x) − (y − 2\sqrt{y}\sqrt{x} + x) = 2\sqrt{y}\sqrt{x} − 2x = 2\sqrt{x}(\sqrt{y} − \sqrt{x}).

This is > 0 because of the assumption y > x and part b. Hence the desired inequality follows.

2. Let a, b be any real numbers. Define max(a, b) and min(a, b) to be the larger and smaller of the two numbers, respectively. (That is, max(a, b) = a if a ≥ b and max(a, b) = b if b ≥ a. Similarly for the minimum.) Show that

max(a, b) = \frac{a + b}{2} + \frac{|a − b|}{2}

and

min(a, b) = \frac{a + b}{2} − \frac{|a − b|}{2}.

Solution: Geometrically, the average \frac{a+b}{2} is the midpoint of the line segment along R between a and b and |a − b| is the distance between a and b (the length of the line segment). So starting at the midpoint and going 1/2 the distance to the right gives the maximum of the endpoints, and going 1/2 the distance to the left gives the minimum of the endpoints. More analytically, we can also prove these by breaking into cases. Suppose first that a ≥ b so a is the right endpoint. Then |a − b| = a − b, and

\frac{a + b}{2} + \frac{|a − b|}{2} = \frac{a + b}{2} + \frac{a − b}{2} = a = max(a, b),

while

\frac{a + b}{2} + \frac{|a − b|}{2} = \frac{a + b}{2} − \frac{a − b}{2} = b = min(a, b).

If b is the maximum and a is the minimum, then

\frac{a + b}{2} + \frac{|a − b|}{2} = \frac{a + b}{2} + \frac{b − a}{2} = b = max(a, b),

while

\frac{a + b}{2} − \frac{|a − b|}{2} = \frac{a + b}{2} − \frac{b − a}{2} = a = min(a, b).

3. Show by mathematical induction that

(1 + x)^{n} ≥ 1 + nx

for all x > −1 and all n ≥ 0. Solution: With n = 0, we have 1 = 1, so the base case is established. Now assume the inequality has been proved for n = k and consider the next case n = k + 1. We have by the induction hypothesis

(1 + x)^{k} ≥ 1 + kx.

Since x > −1, 1 + x > 0, so multiplying this factor on both sides we get

(1 + x)^{k+1} ≥ (1 + kx)(1 + x) = 1 + (k + 1)x + kx^{2} ≥ 1 + (k + 1)x,

since k ≥ 0 and x^{2} ≥ 0. This establishes the desired inequality in all cases.

4. Find a suitable n_{0} and then show by mathematical induction that n! ≥ 5^{n} for all n ≥ n_{0}. Solution: The smallest n_{0} such that n_{0}! ≥ 5^{n_{0}} is n_{0} = 12 since 12! = 479, 001, 600, while 5^{12} = 244, 1140, 625. This gives the base case here: 12! > 5^{12} is true. For the induction step, assume k! > 5^{k} and consider (k + 1)!. We see, by the induction hypothesis (and since k + 1 > 5 whenever k ≥ 12):

(k + 1)! = (k + 1)k! > (k + 1)5^{k} > 5 · 5^{k} = 5^{k+1}.