CHAPTER 3

INTRODUCTORY TOPICS III: MISCELLANEOUS

Answers to Even-Numbered Problems 3.1

√ √ √ √ √ √ √ 2. (a) 2 0 + 2 1 + 2 2 + 2 3 + 2 4 = 2(3 + 2 + 3) (b) (x + 0)2 + (x + 2)2 + (x + 4)2 + (x + 6)2 = 4(x 2 + 6x + 14) (c) a1i b2 + a2i b3 + a3i b4 + · · · + ani bn+1 (d) f (x0 ) x0 + f (x1 ) x1 + f (x2 ) x2 + · · · + f (xm ) xm 2·3+3·5+4·7 6 + 15 + 28 49 4. · 100 = · 100 = · 100 ≈ 144.12 1·3+2·5+3·7 3 + 10 + 21 34 6. (a) The total number of people moving from region i. (b) The total number of people moving to region j .

3.2

2. (a + b)6 = a 6 + 6a 5 b + 15a 4 b2 + 20a 3 b3 + 15a 2 b4 + 6ab5 + b6 . (The coefﬁcients are those in the seventh row of Pascal’s triangle in the text.) 4. (a) = 5 3 = 5·4·3 m 5·4·3·2·1 5! 5! = = = . In general, k 1·2·3 1·2·3·2·1 3! 2! 2! 3! = m(m − 1) · · · (m − k + 1) k!

m! m(m − 1) · · · (m − k + 1) · (m − k)! = . k!(m − k)! (m − k)! k! 8·7·6·5·4 8·7·6 8 8 8 = = = (b) = 56. Also, = 56; 5 3 8−3 1·2·3 1·2·3·4·5 8·7·6·5 9·8·7·6 9 8+1 = = = 126 and = 126. 56 + 4 3+1 1·2·3·4 1·2·3·4 (c)

8 3

+

8 3+1

=

m! m! m m! m m m = = + and = + k+1 k m−k k (m − k)!k! (m − k)!k! (m − k − 1)!(k + 1)! m+1 (m + 1)! m!(k + 1 + m − k) = = = k+1 (m − k)!(k + 1)! (m − k)!(k + 1)!

3.3

2. (a) The total number of units of good i. (b) The total number of units of all goods owned by person j . (c) The total number of units of goods owned by the group as a whole. 4. Because arj − a is independent of the summation index s, it is a common factor when we sum over s, so ¯ m m ¯ ¯ ¯ ¯ s=1 (arj − a)(asj − a) = (arj − a) s=1 (asj − a) for each r. Next, summing over r gives m m m m

r=1 s=1

(arj − a)(asj − a) = ¯ ¯

r=1

(arj − a) ¯

s=1

(asj − a) ¯

(∗∗)

Using the properties of sums and the deﬁnition of aj , we have ¯ m r=1 m m

(arj − a) = ¯

r=1

arj −

r=1

a = maj − ma = m(aj − a) ¯ ¯ ¯ ¯ ¯

m s=1 (asj

Similarly, replacing r with s as the index of summation, one also has Substituting these values into (∗∗) then conﬁrms (∗). © Knut Sydsæter and Peter Hammond 2006

− a) = m(aj − a). ¯ ¯ ¯

CHAPTER 3

INTRODUCTORY TOPICS III: MISCELLANEOUS

9

3.4

2. x = 2. (x = −1, 0, and 1 make the equation meaningless. Multiplying each term by the common denominator x(x − 1)(x + 1) yields 2x(x 2 − 3x + 2) = 0, or 2x(x − 1)(x − 2) = 0. Hence, x = 2 is the only solution.) 4. (a) x ≥ 0 is necessary, but not sufﬁcient. (b) x ≥ 50 is sufﬁcient, but not necessary. (c) x ≥ 4 is necessary and sufﬁcient. √ √ 6. (a) No solutions. (Squaring each side yields x−4 = x+5−18 x + 5+81, which reduces to x + 5 = 5, with solution x = 20. But x = 20 does not satisfy the given equation.) (b) x = 20 √ √ (i) (ii) 8. (a) x+ x + 4 = 2 ⇒ x + 4 = 2−x ⇒ x+4 = 4−4x+x 2 ⇒ x 2 −5x = 0 ⇒ x−5 = 0 ⇐ x = 5. Here implication (i) is incorrect (x 2 − 5x = 0 ⇒ x − 5 = 0 or x = 0.) Implication (ii) is correct, but it breaks the chain of implications. (b) x = 0. (After correcting implication (i), we see that the given √ equation implies x = 5 or x = 0. But only x = 0 is a solution; x = 5 solves x − x + 4 = 2.)

3.5

2. (a) Logically the two statements are equivalent. (b) Appending the second statement is still an expressive poetic reinforcement.

3.6

2. F ∩ B ∩ C is the set of all female biology students in the university choir; M ∩ F the female mathematics students; (M ∩ B) \ C \ T the students who study both mathematics and biology but neither play tennis nor belong to the university choir. 4. (a) B ⊂ M (b) F ∩ B ∩ C = ∅ (c) T ∩ B = ∅ (d) F \ (T ∪ C) ⊂ B

6. (b) and (c) are true, the others are wrong. (Counter example for (a), (d), and (f): A = {1, 2}, B = {1}, C = {1, 3}. As for (e), note in particular that A ∪ B = A ∪ C = A whenever B and C are subsets of A, even if B = C.) 8. (a) Consider Fig. M3.6.8, and let nk denote the number of people in the set marked Sk , for k = 1, 2, . . . , 8. Obviously n1 + n2 + · · · + n8 = 1000. The responses imply that: n1 + n3 + n4 + n7 = 420; n1 + n2 + n5 + n7 = 316; n2 +...