HOMEWORK 1 SOLUTIONS Problem 1 Let F (n) be ”all the polynomials p(x) with degree less than n can be written as (x − b)q(x) + r where degree q(x) is one less than p(x)” F (1): a degree zero polynomial is just an integer which has the desired form with q(x) = 0 F (n) → F (n + 1): without loss of generality we can assume p(x) is a degree n polynomial. p(x) = an xn + an−1 xn−1 + ... + a0 = an xn − an bxn−1 + an bxn−1 + an−1 xn−1 + ... + a0 = an xn−1 (x − b) + (an bxn−1 + an−1 xn−1 + ... + a0 ) by using F (n) for the second term we get: p(x) = an xn−1 (x − b) + (an bxn−1 + an−1 xn−1 + ... + a0 ) = an xn−1 (x − b) + (x − b)q0 (x) + r = (x − b)(an xn−1 + q0 (x)) + r for some q0 (x). So we have proved F (n + 1) with q(x) = an xn−1 + q0 (x). 1. Problem 2 Let Sk (n) be the sum of the first n k-th powers. S4 (n) is the size of the set A := {(i, j, k, l, m) : i, j, k, l, m ∈ N, i, j, k, l ≤ m, m ≤ n} We introduce symmetry by defining sets B := {(i, j, k, l, m) : i, j, k, l, m ∈ N, i, j, k, m ≤ l, l ≤ n} C := {(i, j, k, l, m) : i, j, k, l, m ∈ N, i, j, l, m ≤ k, k ≤ n} D := {(i, j, k, l, m) : i, j, k, l, m ∈ N, i, k, l, m ≤ j, j ≤ n} E := {(i, j, k, l, m) : i, j, k, l, m ∈ N, j, k, l, m ≤ i, i ≤ n} By the principle of inclusion-exclusion, |A ∪ B ∪ C ∪ D ∪ E| 5 5 5 5 5 = |A| − |A ∩ B| + |A ∩ B ∩ C| − |A ∩ B ∩ C ∩ D| + |A ∩ B ∩ C ∩ D ∩ E| 1 2 3 4 5 Date: October 11, 2013. 1 2 HOMEWORK 1 SOLUTIONS Note that A ∪ B ∪ C ∪ D ∪ E = {(i, j, k, l, m) : i, j, k, l, m ∈ N, i, j, k, l, m ≤ n} ⇒|A ∪ B ∪ C ∪ D ∪ E| = n5 A ∩ B = {(i, j, k, l, m) : i, j, k, l, m ∈ N, i, j, k ≤ l, l ≤ n, l = m} ⇒|A ∩ B| = S3 (n) and similarly, |A ∩ B ∩ C| = S2 (n), |A ∩ B ∩ C ∩ D| = S1 (n), |A ∩ B ∩ C ∩ D ∩ E| = S0 (n) = n. So n5 = 5S4 (n) − 10S3 (n) + 10S2 (n) − 5S1 (n) + n which gives us after rearranging, the desired formula 1 5 (n + 10S3 (n) − 10S2 (n) + 5S1 (n) − n) 5 1 1 1 1 1 = n5 + n2 (n + 1)2 − n(n + 1)(2n + 1) + n(n + 1) − n 5 2 3 2 5 1 5 1 4 1 3 1 = n + n + n − n 5 2 3 30 S4 (n) = Let us now prove the above formula by induction. We define the polynomial 1 1 function p(n) = 15 n5 + 12 n4 + 13 n3 − 30 n. For n = 1, p(1) = 15 + 12 + 13 − 30 =1= 5 1 = S4 (1), so the formula holds for n = 1. Now suppose that S4 (k) = p(k) for some k ≥ 1. Then S4 (k + 1) = S4 (k) + (k + 1)4 1 1 1 1 = k 5 + k 4 + k 3 − k + (k + 1)4 5 2 3 30 1 1 1 1 = k 5 + k 4 + k 3 − k + k 4 + 4k 3 + 6k 2 + 4k + 1 5 2 3 30 1 3 13 119 = k 5 + k 4 + k 3 + 6k 2 + k+1 5 2 3 30 and 1 1 1 1 (k + 1)5 + (k + 1)4 + (k + 1)3 − (k + 1) 5 2 3 30 1 5 1 4 3 2 = (k + 5k + 10k + 10k + 5k + 1) + (k 4 + 4k 3 + 6k 2 + 4k + 1) 5 2 1 3 1 2 + (k + 3k + 3k + 1) − (k + 1) 3 30 1 5 3 4 13 3 119 2 = k + k + k + 6k + k+1 5 2 3 30 p(k + 1) = so indeed we have S4 (k + 1) = p(k + 1). By mathematical induction, the formula S4 (n) = holds for all n ∈ N. 1 5 1 4 1 3 1 n + n + n − n 5 2 3 30 HOMEWORK 1 SOLUTIONS 3 Problem 3 Let Sk (n) denote the sum of the First n-th powers as in the notes for lecture 1. Prove by induction that Sk (n) is a polynomial (whose coefficients are rational numbers) of degree k + 1 in n. (Hint: You should prove this by induction on k. You should use as your induction hypothesis that Sj (n) is a polynomial of degree j + 1 for all j smaller than k. [This is sometimes called strong induction.] The last page of the notes for lecture 1 give you a good guess for what the leading term of Sk (n) should be. Express the rest of it as a combination of Sj (n)’s for smaller j.) Let’s prove the statement of the problem for all integers k ≥ 0 by induction in k. To check the base case for k = 0, we have to check, that S0 (n) is a polynomial in n of degree 1. Indeed, S0 (n) = 10 + 20 + · · · + n0 = 1 + 1 + · · · + 1 = n. To do the induction step we start with the guess “the highest term in Sk (n) is nk+1 /(k + 1)” and define Rk (n) to be Sk (n) − nk+1 /(k + 1). It is enough to prove, that Rk (n) is a polynomial in n of degree not greater than k, because then Sk (n) = nk+1 /(k + 1) + Rk (n) is a polynomial of degree k + 1. Note, that from the definition of Sk (n) we have S(0) = 0 and for n ≥ 1 we have Sk (n) = Sk (n − 1) + nk . Using this we compute Rk (0) = Sk (0) − 0k+1 /(k + 1) = 0 and Rk (n)−Rk (n−1) = Sk (n)− (n − 1)k+1 (n − 1)k+1 − nk+1 nk+1 −Sk (n−1)+ = nk + = pk (n). k+1 k+1 k+1 Here pk (n) is a polynomial of degree less or equal than k + 1, so it can be written as pk (n) = nk + (n − 1)k+1 − nk+1 = ak+1 nk+1 + ak nk + ak−1 nk−1 + · · · + a0 . k+1 Note, that we can compute (1) ak+1 = 0 + 1−1 = 0, k+1 ak = 1 + −(k + 1) + 0 = 0. k+1 Therefore pk (n) is a polynomial of degree less or equal than k − 1, so pk (n) = ak−1 nk−1 + ak−2 nk−2 + · · · + a0 . Thus Rk satisfies the following relations: (2) Rk (0) = 0, Rk (n) = Rk (n − 1) + ak−1 nk−1 + ak−2 nk−2 + · · · + a0 . We claim, that this means, that (3) Rk (n) = ak−1 Sk−1 (n) + ak−2 Sk−2 (n) + · · · + a0 S0 (n), so Rk (n) is indeed a polynomial in n of degree less or equal than k, because by induction hypothesis Sj (n) for j < k is a polynomial in n of degree j + 1. We can check equation (3), for example, by induction in n (yes, this is an induction inside 4 HOMEWORK 1 SOLUTIONS the induction step of another induction). We have Rk (0) = 0 = ak−1 0 + ak−2 0 + · · · + a0 0 for the base case n = 0. For induction step we have Rk (n) = Rk (n − 1) + ak−1 nk−1 + ak−2 nk−2 + · · · + a0 = ak−1 Sk−1 (n−1)+ak−2 Sk−2 (n−1)+· · ·+a0 S0 (n−1)+ak−1 nk−1 +ak−2 nk−2 +· · ·+a0 = ak−1 (Sk−1 (n − 1) + nk−1 ) + ak−2 (Sk−2 (n − 1) + nk−2 ) + · · · + a0 (S0 (n − 1) + 1) = ak−1 Sk−1 (n) + ak−2 Sk−2 (n) + · · · + a0 S0 (n) as desired. Alternate solution to Homework 1 Problem 3. Our goal is to show Sk (n) is a polynomial in n by induction on k. The base case . k = 1 is true since we know that S1 (n) = n(n+1) 2 Now we get to assume S1 (n), S2 (n), . . . , Sk−1 (n) are polynomials. We have a k+1 guess that the leading term of Sk (n) is nk+1 . Thus we write a sum which equals the guess by telescoping. n X j k+1 nk+1 (j − 1)k+1 = − k + 1 j=1 k + 1 k+1 n k+1 k+1 X k k−1 l−1 2 l = j − j + · · · + (−1) j k−l+1 + . . . k + 1 k + 1 j=1 k+1 k+1 2 Sk−1 (n) + · · · + (−1)l−1 l Sk−l+1 (n) + . . . k+1 k+1 Now since every term in the equality between the first and fourth lines is known to be a polynomial except for Sk (n), we determine that Sk (n) is a polynomial. = Sk (n) − Problem 4 Sx,y = {p − q | p < x, q > y, p ∈ Q, q ∈ Q}. So, ∀r ∈ Sx,y , r = p − q with p < x, q > y, p ∈ Q, q ∈ Q. =⇒ r = p − q < x − q < x − y. So, x − y is an upper bound of Sx,y and therefore Sx,y is bounded above. Take any real number > 0. ∃N ∈ N such that N > 2 . (why?) =⇒ N 2 > 1. =⇒ N (x − (x − 2 )) > 1 and N ((y + 2 ) − y) > 1. =⇒ N x − N (x − 2 ) > 1 and N (y + 2 ) − N y > 1. =⇒ ∃P, M ∈ N such that N x > M > N (x − 2 ) and N (y + 2 ) > P > N y. (why?) P =⇒ x > M N > x − 2 and y + 2 > N > y. M P M P P =⇒ N − N ∈ Sx,y and N − N > (x − 2 ) − N > (x − 2 ) − (y + 2 ) = x − y − . Therefore, x − y is the least upper bound of Sx,y . (Make sure you understand why this follows from the previous sentence) [Something to think about: We have been using the preposition “the” before “least upper bound”, but is that justified? Can a set of real numbers have more than one least upper bound?] HOMEWORK 1 SOLUTIONS 5 Problem 5 There are many ways to construct examples with the desired properties, and here is one of them. Take x1 = 0.111 · · · 1333 · · · 3 (n 1’s follwed by m−n 3’s) and y1 = 0.111 · · · 1666 · · · 6 (n 1’s follwed by m − n 6’s). Now add an extra digit 5 at the end of the decimal expansion for x1 and y1 to get x2 and y2 , respectively. So x2 = 0.111 · · · 1333 · · · 35, y2 = 0.111 · · · 1666 · · · 65. It is clear tm (x1 ) = x1 = tm (x2 ) and tm (y1 ) = y1 = tm (y2 ) On the other hand, we compute x1 + y1 = 0.222 · · · 22999 · · · 9 (n 2’s follwed by m − n 9’s) x2 + y2 = 0.222 · · · 23 (n − 1 2’s follwed by a 3). Therefore we have tn (x1 +y1 ) = 0.222 · · · 22 (n 2’s) and tn (x2 +y2 ) = 0.222 · · · 23 (n − 1 2’s follwed by a 3). Hence x1 , x2 , y1 and y2 have the desired properties.