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Chapter 9 Generating functions

Chapter 9 Generating functions. Yen-Liang Chen Dept of Information Management National Central University. 9.1. Introductory examples. Ex 9.1. 12 oranges for three children, Grace, Mary, and Frank.

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Chapter 9 Generating functions

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  1. Chapter 9 Generating functions Yen-Liang Chen Dept of Information Management National Central University

  2. 9.1. Introductory examples • Ex 9.1. • 12 oranges for three children, Grace, Mary, and Frank. • Grace gets at least four, and Mary and Frank gets at least two, but Frank gets no more than five. • (x4+ x5+ x6+ x7+ x8) (x2+ x3+x4+ x5+ x6)(x2+ x3+x4+ x5) • The coefficient of x12 is the solution.

  3. Ex 9.2 • Four kinds of jelly beans, Red, Green, White, Black • In how many ways can we select 24 jelly beans so that we have an even number of white beans and at least six black ones? • Red (green): 1+ x1+ x2+….+ x23+ x24 • White: 1+ x2+ x4+….+ x22+ x24 • Black: x6+ x7+….+ x23+ x24 • f(x)=(1+ x1+ x2+….+ x23+ x24)(1+ x2+ x4+….+ x22+ x24)(x6+ x7+….+ x23+ x24) • The coefficient of x24 is the solution.

  4. Ex 9.3. • How many nonnegative integer solutions are there for c1+c2+c3+c4=25? • f(x)=(1+ x1+ x2+….+ x24+ x25)4 • The coefficient of x25 is the solution.

  5. 9.2. Definition and examples: calculational techniques • Ex 9.4. (1+x)n is the generating function for the sequence C(n, 0), C(n, 1),…, C(n, n), 0,0,0…

  6. Ex 9.5 • (1-xn+1)/(1-x) is the generating function for the sequence 1,1,1,…,1, 0, 0,0…, where the first n+1 terms are 1. • 1/(1-x) is the generating function for the sequence 1,1,1,…,1,…. • 1/(1-x)2 is the generating function for the sequence 1,2,3,4,…. • x/(1-x)2 is the generating function for the sequence 0,1,2,3,…. • (x+1)/(1-x)3 is the generating function for the sequence 12,22,32,42,…. • x(x+1)/(1-x)3 is the generating function for the sequence 02,12,22,32,42,….

  7. Ex 9.6 • 1/(1-ax) is the generating function for the sequence a0,a1,a2,a3,…. • Let f(x)=1/(1-x). Then g(x)=f(x)-x2 is the generating function for the sequence 1,1,0, 1, 1,…,…. • Let ai=i2+i for i0. Then its generating function is [x(x+1)/(1-x)3]+[x/(1-x)2]=2x/(1-x)3

  8. Define C(n, r) for nR • Since when nZ+, we have • So for n R we define • For example, if n is positive, we have

  9. Tayor’s and Maclaurin’s Series • f(x)=f(c)+ (x-c) f(c) +(x-c)2f“(c)/2!+ (x-c)3f(3)(c)/3! + (x-c)4f(4)(c)/4! +..+ (x-c)nf(n)(c)/n! • Let c=0. Then we have f(x)=f(0)+ (x) f(0) +(x)2f“(0)/2!+ (x)3f(3)(0)/3! + (x)4f(4)(0)/4! +..+ (x)nf(n)(0)/n!

  10. Ex 9.7

  11. Ex 9.9

  12. Ex 9.10. • Determine the coefficient of x15 in f(x)=(x2+x3+x4+…)4. • (x2+x3+x4+…)=x2(1+x+x2+…)=x2/(1-x) • f(x)=(x2/(1-x))4=x8/(1-x)4 • Hence the solution is the coefficient of x7 in (1-x)-4, which is C(-4, 7)(-1)7=C(10, 7).

  13. Ex 9.11 • In how many ways can we select, with repetition allowed, r objects from n distinct objects? • Consider f(x)=(1+x+x2+…)n • (1/(1-x))n=1/(1-x)n • The coefficient of xr is C(n+r-1, r) • The answer is the coefficient of xr in f(x).

  14. Ex 9.12 • x/(1-x)= x1+ x2+x3+x4+…. • The coefficient of xj in (x1+x2+x3+x4+….)i is the number of ways that we form the integer j by i summands. • The number of ways to form an integer n is the coefficient of xn in the following generating function.

  15. Ex 9.14. • In how many ways can a police captain distribute 24 rifle shells to four police officers, so that each officer gets at least three shells but not more than eight. • f(x)= (x3+x4+ x5+ x6+x7+x8)4 =x12(1+x+x2+x3+x4+x5)4 =x12[(1-x6)/(1-x)]4 • the answer is the coefficient of x12 in (1-x6)4(1-x)-4

  16. Ex 9.17. • How many four element subsets of S={1, 2,…, 15} contains no consecutive integers? • {1, 3, 7, 10}  0, 2, 4, 3, 5 • there exists a one-to-one correspondence between the four-element subsets to be counted and the integer solutions to c1+c2+c3+c4+c5=14 where 0c1, c5 and 2c2, c3, c4. • The answer is the coefficient of x14 in the following formula f(x)=(1+x+x2+x3+…)(x2+x3+x4+…)3(1+x+x2+x3+…)=x6(1-x)-5

  17. Ex 9.17. • {1, 3, 7, 10}  0, 1, 3, 2, 5 • there exists a one-to-one correspondence between the four-element subsets to be counted and the integer solutions to c1+c2+c3+c4+c5=11 where 0c1, c5 and 1c2, c3, c4. • The answer is the coefficient of x14 in the following formula • f(x)=(1+x+x2+x3+…)(x+x2+x3+x4+…)3(1+x+x2+x3+…)=x3(1-x)-5

  18. Ex 9.18. • Brianna takes an examination until she passes it. Suppose in each test the probability of failure is 0.8, and of success is 0.2. • Let Y denote the number of times Brianna expects to take the exam before she passes it. • Please compute E(Y) and E(Y2).

  19. compute E(Y)

  20. compute E(Y2).

  21. the convolution of sequences • Ex 9.19. Let f(x)=x/(1-x)2=0+1x+2x2+3x3+…, where ai=i • g(x)=x(x+1)/(1-x)3=0+12x+22x2+32x3+…, where bi=i2 • h(x)=f(x)g(x)=c0+c1x+c2x2+c3x3+… • ck=a0bk+a1bk-1+a2bk-2+…+ak-2b2+ak-1b1+akb0 • ck=the sequence c is the convolution of sequences a and b

  22. 9.3. Partition of integers • p(x) is the number of partitions for x. • For n, the number of 1’s is 0 or 1 or 2 or 3…. The power series is 1+x+x2+x3+x4+…. • For n, the number of 2’s can be kept tracked by the power series 1+x2+x4+x6+x8+…. • For n, the number of 3’s can be kept tracked by the power series 1+x3+x6+x9+x12+…. • f(x)=(1+x+x2+x3+x4+…)(1+x2+x4+x6+x8+x10+…) (1+x3+x6+x9+…) …(1+x10+…) =1/(1-x)1/(1-x2) 1/(1-x3) … 1/(1-x10) • At last, we have the following series for p(n) by the coefficient of xn

  23. Ex 9.21 • Find the number of ways an advertising agent can purchase n minutes if the time slots come in blocks of 30, 60, 120 seconds. • Let 30 seconds represent one time unit. • a+2b+4c=2n • f(x)= (1+x+x2+x3+x4+…) (1+x2+x4+x6+x8+…)( 1+x4+x8+x12+…) =1/(1-x)  1/(1-x2)  1/(1-x4). • The coefficient of x2n is the answer to the problem.

  24. Examples • Ex 9.22. pd(n) is the number of partitions of a positive integer n into distinct summands. • Pd(x)=(1+x)(1+x2)(1+x3)..… • Ex 9.23. po(n) is the number of partitions of a positive integer n into odd summands. • Po(x)= (1+x+x2+x3+x4+…) (1+x3+x6+x9+x12+…)( 1+x5+x10+x15+…)… • Po(x)=1/(1-x)  1/(1-x3)  1/(1-x5)  1/(1-x7)  ... • Pd(x)= Po(x)

  25. Ex 9.24. • poo(n) is the number of partitions of a positive integer n into odd summands and such summands must occur an odd number of times. • Poo(x)= (1+x+x3+x5+x7+…) (1+x3+x9+x15+…)( 1+x5+x15+x25+…)…

  26. 9.4. The exponential generating function

  27. Ex 9.26 • In how many ways can four of the letters in ENGINE be arranged? • f(x)=[1+x+(x2/2!)]2[1+x]2, and the answer is the coefficient of x4/4!.

  28. important series

  29. Ex 9.28. • We have 48 flags, 12 each of the colors red, white, blue and black. Twelve flags are placed on a vertical pole to show signal. • How many of these use an even number of blue flags and an odd number of black flags? l

  30. Ex 9.28 • how many of these use at least three white flags or no white flag at all?

  31. Ex 9.29. • A company hires 11 new employees, and they will be assigned to four different departments, A, B, C, D. Each department has at least one new employee. In how many ways can these assignments be done?

  32. 9.5. The summation operator • Let f(x)=a0+a1x+a2x2+a3x3+…. Then f(x)/(1-x) generate the sequence of a0, a0+a1, a0+a1+a2, a0+a1+a2+a3,… So we refer to 1/(1-x) as the summation operator.

  33. Ex 9.30. • 1/(1-x) is the generating function for the sequence 1, 1, 1, 1, 1,… • [1/(1-x)] [1/(1-x)] is the generating function for the sequence 1,2,3,4,5,… • x+x2 is the generating function for the sequence 0, 1, 1, 0, 0, 0,… • (x+x2) /(1-x) is the generating function for the sequence 0, 1, 2, 2, 2, 2, … • (x+x2) /(1-x)2 is the generating function for the sequence 0, 1, 3, 5, 7, 9, 11, … • (x+x2) /(1-x)3 is the generating function for the sequence 0, 1, 4, 9, 16, 25, 36, …

  34. Ex 9.31. • g(x)= 1/(1-x)=1+x+x2+x3+x4+… • q(x)=dg(x)/dx=1/(1-x)2=1+2x+3x2+4x3+.… • r(x)=xq(x)=x/(1-x)2 = x+2x2+3x3+4x4+.… • xdr(x)/dx=(1+x)/(1-x)3= x+22x2+32x3+42x4+. • x(1+x)/(1-x)4 = x+(12+22)x2+(12+22+32)x3+(12+22+32+42)x4+.…

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