Illinois State University Mathematics Department

 MAT 409: Topics in Algebra and Combinatorics for K-8 Teachers

Notes on Assignment #12

 Assignment #12 Read Chapter 6 Chapter 7 Chapter 8 Problems Chapter 7 (pp 190-195): 2,3,4,7,8,9,12,17 Chapter 8 (pp 203-205): 1d,2,3,7,9,11

Chapter 7
• #2: Based on the information in the table, f(x) is a quadratic function because a constant difference appears in the second-differences row of the differences table.
•  x 1 2 3 4 5 6 f(x) 12 28 50 78 112 152 D1 ---> 16 22 28 34 40 D2 ---> 6 6 6 6 ..
• #3: We expect the first constant difference to occur in row D3 because the function is a third-degree polynomial. Here we see that the first constant difference is 12.
 x 1 2 3 4 5 6 f(x) 5 26 75 164 305 510 D1 ---> 21 49 89 141 205 . D2 ---> 28 40 52 64 . D3 ---> 12 12 12 .
• #4: (See page 190 in our text for the difference table of the general quadratic, ax2+bx+c.
 Row 1 2 3 4 5 6 Sum 1 5 12 22 35 51 D1 ---> 4 7 10 13 16 . D2 ---> 3 3 3 3 . D3 ---> . . . .
• 2a = 3 --> a = 3/2; 3a + b = 4, a = 3/2 --> b = -1/2; a + b + c = 1, a = 3/2, b = -1/2 --> c = 0. So an explicit row sum representation is S(n) = 3/2x2 - 1/2x.
• #7: See figure (1b) on page 189.
• #8: Sorry, this is a repeat of #2!
• #9: Using difference table for problem #2 and the general quadratic-polynomial difference table (page 190), we have:
• 2a = 6 --> a = 3; 3a + b = 16, a = 3 --> b = 7; a + b + c =12, a = 3, b = 7 --> c = 2.
• Desired representation is 3x2 + 7x + 2.
• #12:
• The next two rows are 6,10,12,12,10,6 and 7,12,15,16,15,12,7.
• The desired function is Sum(row n) = (1/6)n3+(1/2)n2+(1/3)n = (1/6)(n3+3n2+2n) = (1/6)(n)(n+1)(n+2).
• S(100) = 171,700
• #17:
• Row 6: 256, 289, 324, 361, 400, 441; Row 7: 484, 529, 576, 625, 676, 729, 784
• In the row-sums difference table, a constant difference of 30 occurs in row D5, indicating the polynomial that explicitly represents the row sums is a quintic polynomial (5th-degree polynomial).

Chapter 8
• #1d: 2 - x - x3 + 2x4 - x5 + x6 - x7 + ...
• #2:
•  (a) AB AB = 1 + 3x + 3x2 + 2x3 + 4x4 + 5x5 + x6 + x7 (b) AC AC = 1 + x - x3 + x5 + x6 (c) AD AD = 1 + 2x + 3x2 + 3x3 + 3x4 + ... (d) DE DE = 1 + x + x2 + 2x3 + 2x4 + 2x5 + 3x6 + 3x7 +3x8 + ...
• #3: (1 + x + x2 + x3)(1 + x + x2 + x3 + x4 + x5) = 1 + 2x + 3x2 + 4x3 + 4x4 + 4x5 + 3x6 + 2x7 + x8.

• #7: (1 + x + x2 + x3)(1 + x2 + x4)(1 + x + x2 + x3 + ...) = 1 + 2x + 4x2 + 6x3 + 8x4 + 10x5 + 11x6 + 12x7 + 12x8+ ...
• The coefficient of the 7th-powered term is 12, so there are 12 ways to buy 7 pastries
• #9: (1 + x5 + x10 + ... + x100)(1 + x10 + x20 + ... + x100)(1 + x25 + x50 + ... + x100)(1 + x50 + x100)
• After expanding this product and collecting like terms, the coefficient of the 100-degree term will be the number of ways to make the change.
• #11:
• (a) 0 <= k <= 9 (k is between 0 and 9, inclusive)
• (b) 60 sets as described here. You can determine this value by summing the coefficients in the expansion of the generating function shown below.
• (c) (1 + x + x2 + x3)(1 + x + x2 + x3 + x4)(1 + x + x2)
• (d) Revise the generating function to account for the restriction: (x2 + x3)(x + x2 + x3 + x4)(1 + x + x2). Once expanded, look at the coefficient of the 6th-powered term in the expansion. This is the desired response is 6.