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Polynomial Coefficients with Generating Functions

Learn how to calculate coefficients of generating functions and solve combinatorial problems using polynomial expansions and binomial identities. Examples provided for better understanding.

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Polynomial Coefficients with Generating Functions

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  1. MATH 310, FALL 2003(Combinatorial Problem Solving)Lecture 26, Monday, November 3

  2. 6.2. Calculating Coefficients of Generating Functions • Homework (MATH 310#9M): • Read 6.3. • Do 6.2: all odd numberes problems • Turn in 6.2: 2,4,6,14,20,32,38 • Volunteers: • ____________ • ____________ • Problem: 38.

  3. Polynomial Expansions • (1 – xn+1)/(1 - x) = 1 + x + ... + xn. • 1/(1 – x) = 1 + x + x2 + ... + xn + ... • (1 + x)n = 1 + C(n,1)x + C(n,2)x2 + ... + C(n,r)xr + ... + C(n,n)xn. • (1 – xm)n = 1 - C(n,1)xm + C(n,2)x2m + ... + (-1)rC(n,r)xrm + ... + (-1)nC(n,n)xnm. • 1/(1 – x)n = 1 + C(1+n-1,1)x + C(2 + n –1,2)x2 + ... + C(r + n – 1,r)xr + ... • If h(x) = f(x)g(x), where • f(x) = a0 + a1x + a2x2 + ... and • g(x) = b0 + b1x + b2x2 + ..., then • h(x) = a0b0 + (a1b0 + a0b1)x + (a2b0 + a1b1 + a0b2)x2 + ...

  4. Example 1. • Find the coefficient of x16 in (x2 + x3 + x4 + ...)5. Determine the coefficient of xr. • Answer: • a16 = C(6 + 5 – 1,6) = C(10,4). • ar = C(r-10+5-1,r-10) = C(r-6,4).

  5. Example 2 • Use generating functions to find the number of ways to collect $15 from 20 distinct people if each of the first 19 can give a dollar (or nothing) and the 20th can give $1 or $5 (or nothing). • Answer: (1 + x)19(1 + x + x5), • a15 = C(19,15) £ 1 + C(19,14) £ 1 + C(19,10) £ 1 = C(19,4) + C(19,5) + C(19,9).

  6. Example 3 • How many ways are there to distribute 25 identical balls into seven distinct boxes if the first box can have no more than 10 balls but any number can go into each of the other six boxes? • Answer: (1 – x11)(1/(1-x)7). • a0b25 + a11b14 • = 1 £ C(25 + 7 – 1,25) + (-1) £ C(14 + 7 – 1, 14) • = C(31,6) – C(20,6).

  7. Example 4 • How many ways are there to select 25 toys from 7 types of toys with between 2 and 6 of each type? • Answer: • (x2 + x3 + ... + x6)7 = x14(1 + x + x2 + x3 + x4)7 = x14 (1 –x5)7/(1-x)7. • c11 = a0b11 + a5b6 + a10b1 • = 1 £ C(11 + 7 – 1,11) + (-C(7,1)) £ C(6 + 7 – 1,6) + C(7,2) £ C(1 + 7 – 1,1) • = C(17,6) – 7C(12,6) + C(7,2)7.

  8. Example 5 – Binomial Identity • Verify the binomial identity: • C(n,0)2 + C(n,1)2 + ... + C(n,n)2 = C(2n,n). • Answer: • (1 + x)n(1 + x)n = (1 + x)2n. • cn = C(2n,n) • = a0bn + a1bn-1 + ... + an-1b1 + anb0 • = C(n,0)C(n,n) + C(n,1)C(n,n-1) + ... + C(n,n-1)C(n,1) + C(n,n)C(n,0). • Since C(n,r) = C(n,n-r) the result follows.

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