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

‘A‘ Section

1. Let x ∈ [1, 3]. Determine the largest and smallest values of |x − 5|, |x + 5|, and 1/|x^{2} − 25|. Solution: If x ∈ [1, 3], then x − 5 ∈ [−4, −2], so the largest and smallest values of |x − 5| are 4 and 2 respectively. Next, x + 5 ∈ [6, 8], so the largest and smallest values of |x + 5| are 8 and 6 respectively. Finally, |x^{2} − 25| = |x − 5||x + 5|, so the largest and smallest values of 1/|x^{2} − 25| are 1/16 and 1/24 respectively. (Think of the graph y = 1/|x^{2} − 25 to see this.)

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} − 5b^{3})^{6}.

Solution: We have

(a^{2}−5b^{3})^{6} = a^{12}−30a^{10}b^{3}+375a^{8}b^{6}−2500a^{6}b^{9}+9375a^{4}b^{12}−18750a^{2}b^{15}+15625b^{18}.

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

(\frac{x^{4} + 7x}{x^{2}})^{3} .

Solution: Dividing inside the power, we are looking at

(x^{2} + 7x^{−1})^{3}

The only x^{3} term will come from squaring the x^{2} and taking x^{−1} to the first power. So the coefficient is \binom{3}{1} · 7 = 21.

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

(1 + 1)^{n} = 2^{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 such that x ≤ B for all x ∈ A.

Solution: FALSE – the set is bounded above, but not necessarily below. An example would be the open interval (−∞, 0) with B = 0.

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 max(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 min(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 = {x/y | x ∈ A, y ∈ B} is also bounded. Solution: FALSE – Let B = (0, 1) and let A = {1}. The set Q is the set {1/y | y ∈ (0, 1)} which is not bounded above.

4.

a. Let A = [0, 4] ∩ (1, 5). What is a = lub A? What is b = glb A? Are a, b ∈ A? Solution: We have A = (1, 4]. Therefore a = 4, which is in A. But b = 1 is not in A.

b. Let B = {x ∈ R | 0 < x^{2} − 4x + 1 < 4}. What is a = lub B? What is b = glb B? Are a, b ∈ B? Solution: By completing the square or using the quadratic formula, we see B = (2 − \sqrt{7}, 2 − \sqrt{3}) ∪ (2 + \sqrt{3}, 2 + \sqrt{7}). Hence a = 2 + \sqrt{7} and b = 2 − \sqrt{7}. Neither is in B.

c. Let C = {x ∈ Q | x^{2} < 5}. What is a = lub C? What is b = glb C? Are a, b ∈ C? Solution: There are rational numbers arbitrarily close to ±\sqrt{5}, so whose squares are arbitrarily close to 5, but ±\sqrt{5} are not themselves rational numbers. It follows that a = \sqrt{5} and b = −\sqrt{5}, but neither is in C.

‘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^{2} < y^{2}. Solution: Since x > 0, x < y implies x^{2} < xy. Similarly, x < y implies xy < y^{2}. But then transitivity implies x^{2} < y^{2}. Conversely, if x^{2} < y^{2}, then 0 < y^{2} − x^{2} = (y − x)(y + x). Since x, y > 0, the sum x + y > 0. This implies that y − x > 0 too, or equivalently x < y.

c. Show that if 0 < x < y, then \sqrt{y} − \sqrt{x} < \sqrt{y − x}. Solution: By part b, 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^{3} + 2^{3} + ⋯ + n^{3} = (1 + 2 + ⋯ + n)^{2}

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

(1^{3} + 2^{3} + ⋯ + k^{3}) + (k + 1)^{3} = (1 + ⋯ + k)^{2} + (k + 1)^{3}

= (\frac{k(k + 1)}{2})^{2} + (k + 1)^{3}

= \frac{(k + 1)^{2}(k^{2} + 4k + 4)}{4}

= \frac{(k + 1)^{2}(k + 2)^{2}}{4}

= (1 + 2 + ⋯ + (k + 1))^{2}

4. Show by mathematical induction that n! ≥ 2^{n} for all n ≥ 4. Solution: The base case here is n = 4 and 4! = 24 > 16 = 2^{4} is true. For the induction step, assume k! > 2^{k} and consider (k + 1)!. We see, by the induction hypothesis:

(k + 1)! = (k + 1)k! > (k + 1)2^{k} > 2 · 2^{k} = 2^{k+1}.

(The third step is valid since k ≥ 4 implies k + 1 > 2.)