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Unformatted text preview: Generating
Functions In this chapter and the next, we continue our study of enumeration, introducing at this time
i the important concept of the generating function. , The problem of making selections, with repetitions allowed, was studied in Chapter I.
There we sought, for example, the number of integer solutions to the equation cl + c; +
c3 + c4 = 25 where c, 2 0 for all 1 5 i 5 4. With the Principle of Inclusion and Exclusion,
in Chapter 8, we were able to solve a more restricted version of the problem, such as
(:1 + C: + (:3 + C4 : 25 with 0 f c, < 10 for all 1 5i 5 4. If, in addition, we wanted C3 to
be even and C3 to be a multiple of 3, we could apply the results of Chapters 1 and 8 to
several subcases. The power of the generating function rests upon its ability not only to solve the kinds of
problems we have considered so far but also to aid us in new situations where additional
restrictions may be involved. 9.] Introductory Examples EXAMPLE 9.] l Instead of deﬁning a generating function at this point. we shall examine some examples
that motivate the idea. While shopping one Saturday, Mildred buys 12 oranges for her children, Grace, Mary, and
Frank. In how many ways can she distribute the oranges so that Grace gets at least four,
and Mary and Frank get at least two, but Frank gets no more than ﬁve? Table 9.1 lists Table 9.1 G M F G M F
4 3 5 6 2 4
4 4 4 6 3 3
4 5 3 ,6 4 2
4 6 2 7 2 3
5 2 5 7 3 2
5 3 4 8 2 2
5 4 3 5 5 2 415 416 EXAMPLE 9.2 Chapter 9 Generating Functions all the possible distributions. We see that we have all the integer solutions to the equation
cl +63 +03 :12where45c1.2 §c3,and2503 :5. Considering the ﬁrst two cases in this table, we ﬁnd the solutions 4 + 3 + 5 : l2 and 4 +
4 + 4 = 12. Now where in our prior algebraic experiences did anything like this happen?
When multiplying polynomials we add the powers of the variable. and here, when we
multiply the three polynomials, (x4 + x5 + x6 + x7 +x8)(.\‘2 + x3 + x4 + x5 + 166W2 + F +164 + 1'5). 1 two of the ways to obtain .t‘ 3 are as follows: 1) From the pi‘oductx4x3x5. where .r_4 is taken from (x4 + xi + X6 + .r7 + x8), x3 from
(x2 + x3 + x4 + x" + x6), and x3 from (x2 + x3 + )r4 +x*’). 2) From the product .r4x“x4, where the ﬁrst x4 is found in the ﬁrst polynomial, the
second x4 in the second polynomial, and the third x4 in the third polynomial. Examining the product
(x4 + x5 + x6 + x7 + 1C8)(x2 + x3 +x4 4— x5 + Jc6)(,r2 + x3 + 1c"l + x5) more closely, we realize that we obtain the product xix/Ix" for every triple (i, j, k) that
appears in Table 9.1. Consequently, the coefﬁcient of x12 in f(x) = (xJr —l— x5 + x6 + x7 + ,r8)(.r2 +x3 + x4 + x5 + ,rf’)(,r2 + x3 + .r‘1 + 2?) counts the number of distributions—namely, 14—that we seek. The function f(x) is
called a generating function for the distributions. But where did the factors in this product come from? The factor x4 + x5 + x6 + x7 + x8. for example, indicates that we can give Grace 4 or
5 0r 6 or 7 or 8 of the oranges. Once again we make use of the interplay between the exclusive
or and ordinary addition, The coefﬁcient of each power of x is 1 because, considering the
oranges as identical objects, there is only one way to give Grace four oranges, one way
to give her ﬁve oranges, and so on. Since Mary and Frank must each receive at least two
oranges, the other terms (xi + x3 + x4 + x5 + x6) and (x2 + x3 + x4 + x5) start with .rZ,
and for Frank we stop at P so that he doesn’t receive more than ﬁve oranges. (Why does
the term for Mary stop at .ré?) Most of us are reasonably convinced now that the coefﬁcient of .r '3 in f(x) yields the
answer. Some, however, may be a bit skeptical about this new idea. It seems that we could
list the cases in Table 9.1 faster than we could multiply out the three factors in f(x) or
calculate the coefﬁcient r” in f (x). At present that may seem true. But as we progress to
problems with more unknowns and larger quantities to distribute, the generating function
will more than demonstrate its worth. (The reader may realize that the rook polynomials of Chapter 8 are examples of generating functions.) For now we consider two more examples. _——_—I———_—___—_ If there is an unlimited number (or at least 24 of each color) of red, green, white, and black
jelly beans, in how many ways can Douglas select 24 of these} candies so that he has an
even number of white beans and at least six black ones? The polynomials associated with the jelly bean colors are as follows: 0 red (green): 1 + x + x3 +  ~ v + x24, where the leading 1 is for 1x0, because one
possibility for the red (and green) jelly beans is that none of that color is selected (1+x3+x4+x6+~~+x24) (x6+x7+x8+_u+x24) 0 white:
0 black: EXAMPLE 9.3 417 9.] Introductory Examples So the answer to the problem is the coefﬁcient of 1‘24 in the generating function
,/‘(x) = (1+x +x2 + a . ‘ +x34)3(i +x2 +x‘ + ~  ~ +x24)(x6 +x7 + x  ~ + x34). One such selection is ﬁve red, three green, eight white, and eight blackjelly beans. This
arises from x3 in the ﬁrst factor. x3 in the second factor, and x8 in the last two factors. _——'—_l—_————_— One more example before closing this section! How many integer solutions are there for the equation 6, + C3 + C3 + 64 = 25 if() f c; for
all 1 5i 5 4‘? We can alternatively ask in how many ways 25 (identical) pennies can be distributed
among four children. For each child the possibilities can be described by the polynomial l + x + x2 + x3 —l—
. >  + x 35. Then the answer to this problem is the coefﬁcient of x35 in the generating function f(x) : (1 +x +x2 + .  . +.t3>)4.
The answer can also be obtained as the coefﬁcient of x25 in the generating function gtx) =(1+x+x3+x3+.~+x25+x26+...)4, , if we rephrase the question in terms of distributing, from a large (or unlimited) number of
pennies, 25 pennies among four children. [Whereas f(x) is a polynomial, got) is apower
series in All Note that the terms .r", for all k 3 26, are never used. So why bother with them?
Because there will be times when it is easier to compute with a power series than with a polynomial. W 1. For each of the following, determine a generating function
and indicate the coefﬁcient in the function that is needed to solve
the problem. (Give both the polynomial and power series forms
of the generating function, wherever appropriate.) Find the number of integer solutions for the following equa—
trons: a) Ci+C2—C3+C4:20.0§C,E7f0f€1lllfif4 b) Cl 62 C3 + c4 — 20. O 5 C, for all 1 5 i 5 4, with c;
and 63 even i
C) C1 Cz+C3lC4 C5T30,256154aﬂd3:Ci58 forallZgigS d) C1+C2+C3+C4’“C5 =30,0:c, for all l Si :5,
with c; even and C3 odd 2. Determine the generating function for the number of ways
to distribute 35 pennies (from an unlimited supply) among ﬁve
children if (a) there are no restrictions; (b) each child gets at
least M: (0) each child gets at least 2¢; (d) the oldest child gets
at least 10¢; and (e) the two youngest children must each get at
least 10¢. 3. 3) Find the generating function for the number of ways to
select 10 candy bars from large supplies of six different kinds. b) Find the generating function for the number of ways to
select, with repetitions allowed, r objects from a collection
ofn distinct objects. 4. a) Explain why the generating function for the number of
ways to have 11 cents in pennies and nickels is (1+x +x2+x3 +~)(l +x5+x10+<~). b) Find the generating function for the number of ways to
have )1 cents in pennies, nickels, and dimes. 5. Find the generating function for the number of integer
solutions to the equation c1 + CZ + C3 + c4 = 20 where
~3 :61, —3563, ~5 5c; 55,and0§c4. 6. For S I la, b, c}, consider the function
we = (1+ ax)(l + bx)(1+ cx)
: l +ax +bx—l—cx +abx2 +acx2
+ bcx2 + (1170?.
Here, in f(x) 0 The coefﬁcient of x0 is 1 — for the subset 0 of S. 0 The coefﬁcient of x1 is a + b + c~for the subsets {a}.
{b}, and {c} of S. 0 The coefﬁcient of x2 is ab +ac + lac—for the subsets
{(1.17}, {a, c}, and {Z}, c} of S, 418 Chapter 9 Generating Functions 0 The coefﬁcient of x3 is abc—for the subset (a, b, C} = S. 3) Give the generating function for the subsets of
Consequently, f (x) is the generating function for the sub— 5 = {as be Q ~ A « a "a 5: ti
SGIS 0f 54 For When we caICUIate f (1), W6 Obtain a sum Wherei“ b) Answer part (a) for selections wherein each element can
each of the eight summands corresponds with a subset of S; be rejected or selected as mam, as thme times‘ the summand 1 corresponds with (5. [If we go one step further
and set a = b = c = l in f(x), then f(1) = 8, the number of
subsets of S.] 9.2
Definition and Examples:
Calculational Techniques In this section we shall examine a number of formulas and examples dealing with power
series. These will be used to obtain the coefﬁcients of particular terms in a generating function.
We start with the following concept. Deﬁnition 9.] Let ao, a1. a2, . t . be a sequence of real numbers. The function
[x .
f(x) : no —— tax + (12x2 +    = Z citxi
1:0 is called the generating function for the given sequence. Where could this idea have come from? , 4"
EXAMPLE 9.4 For “my ” E Z *
(1+x)” = (n) + (">x + (n>x2 +    + (")x”,
I 0 l 2 n so (i + x)" is the generating function fOr the sequence (so(zinwcimnu m. (1—xnt‘)=(i—x)(1+x+x3+x3+m+x”). So
1_xn+l
=1+x+x2++x”,
l—x
and (1 — x”+‘)/(1  x) is the generating function for the sequence 1, l, l, . . . , 1, 0, 0, 0, . . . , where the ﬁrst n + 1 terms are 1‘ 92 Definition and Examples: Calculational Techniques 419 b) Extending the idea in part (a), we ﬁnd that
l :(1—x)(l+x+x2+x3+x4+...)’ SO l
l — x
is the generating function for the sequence 1, l, l, l, . . . . [Note that l/(l — x) = l +
x + x2 + x3 + a   is valid for all real x where [x] < l; it is for this set of values that the geometric series I + x + x: + x3 +    converges. In our work with generating
functions we shall be primarily concerned with the coefﬁcients of the powers of x.
However, later in Example 9.18, we shall use this and two other related series to
evaluate inﬁnite sums for values within the set of values for which each such inﬁnite
series converges] c) With
1 00 _
l_x =l+x+x2+x3+=§x‘,
taking the derivative yields
d l d l
_ l . ~l V I 1 ‘2 I a.
dxl_x dx( x) ( )( x) ( ) (l_x)2
0' ~ 7
: d—(1+x+x2+x"+~) = l+2x+3x"+4x3+~«
x
Consequently,
1
(1—K)2
is the generating function for the sequence 1, 2, 3, 4, . . . , while
’ a .2 3 4
(1 “XV —O+lx+2x +3x +4x +~~ is the generating function for the sequence 0, l, 2, 3, . . . . (1) Continuing from part (c), d x d
_ :_0 ‘22 33
dx(1_x)2 dx( +x+ x + x + ) or
+ l
{Tb—3 21+22x+31x2+42x3+m.
Hence,
)5 + l (1 —x)3
generates 12, 22, 32, . . . , and x(x +1) (1 —X)3 generates 02, 12, 22, 32, . . . . 420 Chapter 9 Generating Functions e) Now let us take one more look at the results in parts (b), (c), ((1) —along with some
extensions. But this time we have a change in the notation: :1+,t~+x3+x3+~ foOC) : l—x . ‘W 61 _
frU) # Xﬁfoﬂ’c) * (l—x)2
ZO+X+2x2+3.r3+‘
. _ d . _ x2+x
fz(A)VX;l;fr(X) * (l_x)3
= 02 +12x + 22x2 + 32x3 +  ~
d x3+4x2+x
f3(X) defzbc) (1_x)4 =03+ l3x+23x2+33x3+~ x4+ llx3+ llx3+x
(lﬂc)5 = 04 + 14x + 24);: + 34163 + ‘ H d
f4(x) ‘xd f3(x) —
x Now look at the output for the Maple code in Fig. 9.1. Here we ﬁnd the numerators for
f0(x), f1(x), . . . , f4 (x), along with those for f5 (x) and f6(x) [where the denominators
are (l — x)6 and (l — x)7, respectively]. The coefﬁcients for these numerators are exactly
the Eulerian numbers we introduced in Example 4.21. We choose not to pursue this here,
but the interested reader, who wants to examine this connection further, should look into
reference [4]. i> f0(x) := 1/(1X); f0(x) 2: l‘x > for i from 1 to 6 do > in(x) := simplify(x*diff(f[(i—l)(x),x)): > print(sox1:(expand((—1)"(i+1)*numer(fl i(x)))));
> 0d: I x z
X +X 7
x3+4x’+x x4+11x3+11x2+x x5+26x4+66x3+26x2+x _ 16+57x5+302x4+302x3+57x2+x Figure 9.1 9.2 Deﬁnition and Examples: Calculational Techniques 42] EXAMPLE 9 6 a) Rewriting the result in part (b) of Example 9.5, we have Upon substituting 2x for y, we then learn that 1 2 =l+(2x)+(2x)2+(2x)3+~~=l+2x+22x2+23x3+~,
— x 7 so 1/(1 — 2x) is the generating function for the sequence 1 (I 20), 2 (2 2‘), 2‘,
23. . . ..Infact,foreacha E R, itfollows that l/(l — ax) = l + (ax) + (ax)2 + (ax)3 +    = l + ax + azx2 + a3x3 +    , so l/(l w ax) is the generating function for
the sequence 1 (= (10), a (= a1), a2, a3, . . . . [Here we want 00 = l for the case where
a = 0.] b) Again, from part (b) of Example 9.5, we know that the generating function for the
sequence 1, l, l, l, . . . is f(x) = l/(l — x). Therefore the function 1 7
_x— 806) = for) “x2 = l—x is the generating function for the sequence 1, l, O, l, l, l, . . . , while the function hm : f(.r)+2x3 = 1 1 generates the sequence 1, l, l, 3, l, l, . . . .
c) Finally, can we use the results of Example 9.5 to ﬁnd a generating function for the
sequence 0, 2, 6, 12, 20, 30, 42, . . . ?
Here we observe that
a():0:OZ+0, a1:2:l2+l,
a2:6=22+2, a3=12:33+3,
04:20:42+4,.... In general, we have (1,1 = n2 + n, for each n 3 0. ,
Using the results from parts (0) and (d) of Example 9.5, we now ﬁnd that x(x+l)+ x *x(x+l)+x(l~x) _ 2x
(1—103 (1—IC)2 (1—03 (l—x)3 is the generating function for the given sequence. (The solution here depends upon
our ability to recognize each an as the sum of n2 and n. If we do not see this, we may
be unable to answer the given question. Consequently, in Example 10.6 of the next
chapter, we shall examine another technique to help us recognize the formula for an .) ————————_—_——_____ For each n e Z+, the binomial theorem tells us that (1 + x)” = + + (9x2 +
   + fax". We want to extend this idea to cases where (a) n < 0 and (b) n is not necessarily an integer. 422 Chapter 9 Generating Functions ‘ with“ “ﬂeet 3:2:r»>'0,2weihava 1 n, (,2 / '_,i<z'(iz’—iit>i<:jzg , ‘ ' " it‘li‘ltn 4r)! i _ .v f', C ’ “a: ; a _ v
IfnjeRiwe’usef’q l ‘ ‘ : ~ —j~ 1*)112352)" . <;
has. thcdeﬁniﬁenpf (31).)": .» , a ‘
’ﬂ’E/Z‘E'Wéhév Thehtiféfeaaﬁiﬁe “ 1C; 5 111.9532; 4 ' way, + <  . ,n .‘ .
EXAMPLE 9.7 For n E Z , the Maclaurin series expansmn for (l + 1:) Is given by (1 l x)‘” — 1+( n)x+( n)( n [Ml/2!
+ (~n)(—n — l)(—n — 2)x3/3! + . .
~1+ Z (—n)(—n — l)(—rz ~— 2) ~ O  (—n — r + UK
—1 M _ r!
x — 1 = Z<—1>”(” J” r:0 r Hence (1 + x)_” = (—0") + (3”)x + (71))? +  « . = :0 (—”)x’. This generalizes the u r binomial theorem of Chapter 1 and shows us that (1 +x)“" is the generating function
for the sequence (“0”), (—1”), (3"), . . . . Find the coefﬁcient of x5 in (1 — 2x)”. 1
EXAMPLE 9'8 With y = —2x, use the result in Example 9.7 to write (1 — 2x)“7 = (1+ y)‘7 =
:10 (77)y’ = :0 (77)(—2x)’. Consequently, the coefﬁcient of x5 is (—57)(——2)5 = (—1)5(7+§— 1)(—32) = (32mg) 2 14,784. For each real number n, the Maclaurin series expansion for (1 + x)” is EXAMPLE 9.9
1+ nx + n(n —1)x2/21+ n(n —1)(n — 2)x3/3! +  ‘ w 00n(n—1)(n——2)~(n—r+l)r
—1+; r! x. EXAMPLE 9.10 EXAMPLE 9.“ l EXAMPLE 9.12 l 423 9.2 Definition and Examples: Calculational Techniques Therefore, (1 + 3104/3 : 1 + Z («1/3><~4/3)(~7/3)  < ‘ ((—3r + 2>/3> r=l
21+ (*l)(—4)(—7) ‘ ' ‘ (—3r + 2%,,
r:l (3x)r r! rl and (1+ 3,’c)‘l/3 generates the sequence 1, —l, (~l)(~4)/2l. ('~l)(—4)(\—7)/3l, . . .,
(—l)(—4)(—7)  ‘  (3r + 2)/r!, . a .. ————————_—___________ Determine the coefﬁcient of x15 in f(.r) = (x2 + x3 + x4 +  v 04‘ Since ()62 +163 +264 + ~  ‘) = x2(l +x +x2 + ~  ) = XZ/(l — x), the coefﬁcient of
x15 in for) is the coefﬁcient of x15 in (xZ/(l — .r))4 : xg/(l — M4. Hence the coefﬁcient
sought is that ofx7 in (1 — x)"4, namely, (7‘)(~1)7 : (—1)7(4+;‘1)(—l)7 = (170) z 120. In general, for n e ZT, the coefﬁcient of x” in f(x) is 0, when 0 5 n 5 7‘ For all n 3 8,
the coefﬁcient of x” in f(x) is the coefﬁcient of x”_8 in (l ~x)‘4, which is (":48) 
(—1)“; 2 (2:33} —————————_—_____ Before continuing, we collect the identities shown in Table 9.2 (on page 424) for future
reference. The next two examples show how generating functions can be applied to derive some
of our earlier results. In how many ways can we select, with repetitions allowed, r objects from n distinct objects? For each of the n distinct objects, the geometric series 1 + x + x2 + x3 + ‘   represents
the possible choices for that object (namely none, one, two, . . .) . Considering all of the n
distinct objects, the generating function is foe) =(1+x+x2+x3+~)", and the required answer is the coefﬁcient of x’ in f (x). Now from identities 5 and 8 in
Table 9.2 we have ‘ l n l 00 n+i~l r
1 2 3 ...”= : _..__..: J1!
(“H H + ) (1—x> (l—x)" 2( i )‘ i=0 so the coefﬁcient of x’ is the result We found in Chapter 1. W Once again we consider the problem of counting the compositions of a positive integer
n — this time using generating functions. Start with x =x+x2+x3+x4+'~ l—x 424 Chapter 9 Generating Functions Table 9.2 "in a six)“ S=‘ (3) (9x 4: (W + ’ 4+ (axe:1 : . .
i “an wee(31* (riax‘ne (%)a2x?+ than ,
i 3) ea— e (a) + in“ +~(3)x2t + (i: xmﬁ‘ * ‘
* .4)“ («tears/<1: ~.,x.>,.~—~;1 ﬂ’+ x? "M". ' j ’ 71;
5) C1y(:1ji;x)‘:g{1,+.x 12x24” x3,+ .p.‘,.f:‘ ; .
é)Elf/{ism}:mice)etaxﬁejfraee;, ' " where, for example, the coefﬁcient of x4 is 1, for the one—summand composition of 4w
namely, 4, To obtain the number of compositions of n where there are two summands,
we need the coefﬁcient ofx” in (x +x2 + x3 +x4 +   )2 = [x/(l — X)l2 = xZ/(l — x)2,
Here, for instance, we obtain x4 in (x + x2 + x3 + x4 +  < )2 from the products x1 ~ x3,
x2 x2, and x3 xl. So the coefﬁcient of x4 in xZ/(l — x)2 is 3—for the three two
summand compositions l + 3, 2 + 2, and 3 + 1 (of 4). Continuing with the threesummand
compositions we now examine (x + )c2 —l— x3 + x4 —l—  « 3 : [x/(l — x)]3 2 x3/(1— x)3.
Once again we look at the ways x4 comes about— namely, from the products x1 ‘ x1 ~ x2,
x1 ~ x2 < x1, x2  xl  x1. So here the coefﬁcient of x4 is 3, which accounts for the composi
tions 1 + 1 —l— 2, l + 2 + 1, and 2 + 1 + l (of 4). Finally, the coefﬁcient of):4 in (x + x2 +
x3 + x4 +~)*1 = [x/(l — x)]4 = x4/(l — x)4 is l—for the one foursummand compo—
sition1+ 1+1+1(0f4), The results in the previous paragraph tell us that the coefﬁcient ofx4 in 1 [x /(1 — x)]i
is l + 3 + 3 + 1 = 8 (= 23), the number of compositions of 4. In fact, this is also the
coefﬁcient of x4 in Zillx/ (1 — x)]i. Generalizing the situation we ﬁnd that the number
of compositions of a positive integer n is the coefﬁcient of x” in the generating function 9.2 Definition and Examples: Calculational Techniques 425 f(.r) = 2:1[x/(l — Ml". But if we set y = x/(l — x), it then follows that Iii/“4:” “(1 it) ‘(1 il—i—i:>l H (l l l
X/(1 * Zr) =.r[1+(2x)+(2x)2 + (2,03 + . . .]
= 20x + Z‘x2 —l— 22x3 + 23x4 + . . . _ So the number of compositions ofa positive integer n is the coefﬁcient ofx” in f(x) — and
this is 2”“‘ (as we found earlier in Examples 1.37. 3.11, and 4.12.) Ml Before we look at any speciﬁc compositions, let us start by examining identity 4 in Table
' 9.2. When x is replaced by 2 in this identity, the result tells us that for all 11 E Z+, 1 +
2 —l— 22 +  . . + 2" : (1 ~ 2”“)/(1 — 2) : 2”“ — 1. [This result was also established by
the Principle of Mathematical Inductionwin part (a) of Exercise 2 for Section 4.1.] All
well and good~but where would one ever use such a formula? In Table 9.3 we ﬁnd the
special compositions of 6 and 7 that read the same left to right as right to left. These are the
palindromes of6 and 7. We ﬁnd that for7 there are 1 + (1 + 2 + 4) = l + (l —— 21 + 22) =
l —l— (23 — l) = 23 palindromes. There is one palindrome with one summand~namely, 7.
There is also one palindrome Where the center summand is 5 and where we place the one
composition of 1 on either side of this summand. TableQJ 1) 6 (I) l)‘ 7 (1)
2) 1+4+1 (l) 2) l+5+1 (1)
3 2+2+2 3 2+3 2
) <2) ) + (2)
4) 1+1+2+1+1 4) 1+1+3+1+1
5) 3+3 5) 3+l+3
6) l+2+2+l 6) 1+2+1+2+1 (4) (4)
7) 2+1+1+2 7) 2+i+1+1+2
8)l+l+1+l+1+l 1’8) i+1+1+1+i+i+1 ‘ J For the center summand 3 we place one of the two compositions of 2 on the right (of 3)
and then match it on the left, with the same composition, in reverse order. This procedure
provides the third and fourth palindromes of 7 in the table. Finally, when the center summand
is 1, we put a given composition of 3 on the right of this 1 and match it on the left with the
same composition, in reverse order. There are 23‘1 = 4 compositions of 3, so this procedure
results in the last four palindromes of 7 in the table. The situation is similar for the palindromes of 6 except for the case where, instead of 0 as the center summand. a plus sign appears in the center. Here we obtain the last 23“1 = 4
palindromes of 6 in the table—one for each composition of 3. Summarizing for n = 6 we
have
i) Center summand 6 1 palindrome
ii) Center summand 4 1 (I 21") palindrome
iii) Center summand 2 2 (= 22‘1) palindromes iv) Plus sign at the center 4 (= 23—1) palindromes 426 Chapter 9 Generating Functions So there are 1+ (1 + 21 + 23) = l + (23 — l) 2 23 palindromes for 6. Now we look at the general situation. For n 2 1 there is one palindrome. If n : 2k + l,
for k e Z+, then there is one palindrome with center summand n. For 1 5 t 5 k, there
are 2’” palindromes of n with center summand n — 2t. (One palindrome for each of the
2’” compositions of t.) Hence the total number of palindromes of n is l + (l + 2l +
22 +  ~  + 2"”) = 1+ (2" — l) : 2" = 2(”"”/2. Now consider n even, say n = 2k, for
k E Z+. Here there is also one palindrome with center summand n and, for l f s f k + 1,
there are 25‘[ palindromes of n with center summand n — 25. (One palindrome for each
of the 2*1 compositions of S.) In addition, there are 2"” palindromes where a plus sign
is at the center. (One palindrome for each of the 2’"1 compositions of k.) In total, n has
1+ (1 + 21+ 22 + 3  a + 2’"2 + 2"”) = 1+ (2" — 1) I 2" = 2’”2 palindromes. The preceding results can be simpliﬁed. Observe that for n 6 2+, n has 2W2J palin—
dromes. Having dealt with compositions (once again)“and palindromes, we continue at this point
with some additional examples dealing with generating functions. In how many ways can a police captain distribute 24 riﬁe shells to four police ofﬁcers so
' that each ofﬁcer gets at least three shells, but not more than eight?
The choices for the number of shells each ofﬁcer receives are given by x3 + x"L + ~  e +
x8. There are four ofﬁcers, so the resulting generating function is f(x) = (x3 +x4 + ~   +x8)4_ We seek the coefﬁcient ofx24 in f(x). With (x3 + x4 + < .  + r8)4 : x13(l + + x2 +
~  +x5)4 = xlz((l — x6)/(1 — x))4, the answer is the coefficient of x” in (l — x6)4 
(1— “‘4 =11 — (W6 + (3%12 " (9)618 + 1524] ii?) + (34%”) + (3419“: + ‘ ' ‘lv
Whichis liii)<“1)12 ‘ (ili24)(—U6 + (3mm 2 l 1’?) * (3(2) +191 : 125 Verify that for all n e z+, (2,7) = 2;, (7)2.
Since (1 + x)2” = [(1 + x)"]2, by comparison of coefﬁcients (of like powers of x),
the coefﬁcient of x” in (1+x)2”, which is (2”), must equal the coefﬁcient of x” in fl [(3) + (ax + (9162+ ‘   + (:W and this is (on) + (mm + (3)912) +    +
. With (it) 2 (air), for 21110 E r E n, the result follows. . . . 1
Determine the coefﬁcrent of x8 in EXAMPLE 9.16 (x ~— 3)(x — 2)2'
Sinai 1/06 ~ a)=(—1/a)(1/(1—(x/a))) =(—l/a)[l+(x/a)+(Jc/Ct)2 + ~ ~ 1 for
any a 75 0, we could solve this problem by ﬁnding the coefﬁcient of x8 in
1/[(x — 3)(x — 2)2] expressed as (—1/3)[1 + (x/3) + (X/3)2 + ~ ~ ~](1/4) [(32) +
(ﬂex/2) + (ﬂex/2V + a  ]' An alternative technique uses the partial fraction decomposition: i _ A + B + C
()c—3)()c—2)2 x—3 x—Z (Jr—2)?
This decomposition implies that 1: A(x —2)2+B(x —2)(x —3)+C(x ~3), EXAMPLE 9.17 427 9.2 Definition and Examples: Calculational Techniques 01‘
0x2+ 0x+l=1=(A+B)x2+(—4A—SB+C)x+(4A+6B—3C). By comparing coefﬁcients (for x2, x, and 1, respectively), we ﬁnd that A + B = 0,
—4A — SB + C 2 O, and 4A + 63 — 3C = 1. Solving these equations yields A = 1, B =
—1, and C = —1. Hence 1 _ 1 1 1
x—2 (x——2)2 (x H W — 2P _ (31> 1— (Ix/31 + 1—ix/2i +01) <1—<:/2)>2
=(e‘>:(2>‘+e>:e>’ 7
+ (2—1) {(32) if?) (a) +6) (if)1} The coefﬁcient Ofxg is (~1/3)(1/3)8 + (1/2)(1/2)8 +(—1/4)(;§)(w1/2)8 =
— [(1/3)9 + 7(1/2)‘0]. x—3 Use generating functions to determine how many four—element subsets of S = {1, 2, 3, . . . ,
15} contain no consecutive integers. 3) Consider one such subset (say {1, 3, 7, 10}), and write 1 5 1 < 3 < 7 < 10 5 15. We
see that this set of inequalities determines the differences 1 ~ 1 = O, 3 — l 2 2, 7 ~
3 = 4, 10 — 7 = 3, and 15 — 10 = 5, and these differences sum to 14. Considering
another such subsetﬁsay {2, 5, 11, 15}, we write 1 5 2 < 5 < 11 < 15 5 15; these
inequalities yield the differences 1, 3, 6, 4, and O, which also sum to 14. Turning things around, we ﬁnd that the nonnegative integers 0, 2, 3, 2, and 7 sum to
14 and they are the differences that arise from the inequalities 1 5 1 < 3 < 6 < 8 5 15
(for the subset {1, 3, 6, 8}). These examples suggest a onetoone correspondence between the fourelement
subsets to be counted and the integer solutions to C; + cz + C3 + C4 + cs = 14 where
O 5 c1, C5, and 2 5 Q, C3, C4. (Note: Here c2, C3, C4 3 2 guarantee that there are no
consecutive integers in the subset.) The answer is the coefﬁcient of x14 in f(x) =(1+x+x2+x3+)(x2+x3+x4+i)3(1+x+x2+x3+~i) = x6(1—— x)_5. This then is the coefﬁcient of x8 in (1 —x)5, which is (85)(—1)8 = (“3* 1) =
(1,2) = 495.
b) Another way to look at the problem is as follows. For the subset {1, 3, 7, 10}, we examine the strict inequalities 0 < l < 3 < 7 <
10 < 16 and consider how many integers there are strictly between each successive
pair of these numbers. Here we get 0, l, 3, 2, and 5: 0 because there is no integer
between 0 and 1, 1 for the integer 2 between 1 and 3, 3 for the integers 4, 5, 6 between
3 and 7, and so on. These five integers sum to 11. When we do the same thing for the
subset {2, 5, ll, 15}, the strict inequalities O < 2 < 5 < 11 < 15 < 16 yield the results
1, 2, 5, 3, and 0, which also sum to 11. 428 EXAMPLE 9.181 I ChaPter 9 Generating Functions On the other hand, we ﬁnd that the nonnegative integers 0, l, 2, l, and 7 add up to
ll and they arise as the numbers of distinct integers between the integers in the ﬁve
successive strict inequalities 0 < l < 3 < 6 < 8 < 16. These correspond to the subset
{ l. 3, 6, 8}. These results suggest a one—to—one correspondence between the desired subsets
and the integer solutions to bi + b; + [93 + b; + [95 = ll, where 0 _<_ b1, b5 and l 5
b2, [93, [74. (Note: In this case, [21, b3. b3 3 l guarantee that there are no consecutive
integers in the subset.) The number of these solutions is the coefﬁcient of x ” in g(x) =(1+x+x2+<~)(x+x3+x3+~~)3(l+x+x2+~) =.r3(l —x)“5. The answer is (—85) (— l)8 = 495, as above. (The reader may now wish to look back
at Supplementary Exercise 15 in Chapter 3.) encountered the idea of the sample space. But now that we know about generating functions
we will be able to deal with a sample space that is discrete but not ﬁnite — that is, a countany
inﬁnite+ sample space. a) Suppose that Brianna takes an actuarial examination until she passes it, Further, sup
pose the probability that Brianna passes the examination on any given attempt is 0.8
and that the result of each attempt, after the first, is .independent of any previous at~
tempt. If we let Pdenote “pass” and F denote “fail”. for any given attempt, then here our
sample space may be expressed as 3] 2 (P, FP, FFP, FFFP, . . .}, where, for example,
Pr (FFP) — the probability Brianna fails the exam twice before she passes it ~ is given
by (0.2)Z(0.8). In addition, the sum of the probabilities for the outcomes in 9’ is (0.8) +
(0.2)(0.8) + (0.2)3(0.8) + (0.2)3(0.8) + a   = :O(O.Z)i(0.8) = (0.8) 2:0(02ﬂ
= (0.8) (I 102) = (0.8) I l, as it should be —— for according to the second axiom of probability (in Section 3.5) we expect Pr(5f) = 1. [Note that S_l‘19':()(0.2)i = I 10.2
follows from the result in part (b) of Example 9.5. The given geometric series con— verges to 1_10_2 because {0.2! < 1.] b) Now suppose we want to know the probability Brianna passes the exam on an even
numbered attempt. That is, we want Pr(A) where A is the event {FR FFFP, . . .}. At this point let us introduce the discrete random variable Y where Y counts the num
ber of attempts up to and including the one where Brianna passes the exam. Then the
probability distribution for Y is given by Pr(Y = y) = (0.2)y’1(0.8), y 3 1. So Pr(A)
can be determined as follows: Pr(A) : {:1 Pr(Y = 21') = :1(0.2)2i‘1(0.8) =
(0.8) 2,240.2)2H : 0.8[(0.2) + (0.2)3 + (0.2)5 + .  .1 = (O.8)(0.2)[1+(O.2)2 +
(0.2)4 +  ~ .1 = (0.8)(0.2) I _ (‘W = = 61. And once again we have used the 0.96
result in part (b) ofExample 9.5, this time withx = (0.2)2, where [(0.2)21 = {0.04} < 1. TThe reader can learn more about countany inﬁnite sets from the material in Appendix 3. iThis example uses material from the optional sections of Chapter 3. It may be skipped without any loss of
continuity. 9.2 Definition and Examples: Calculational Techniques 429 c) Continuing with Y, now we’d like to ﬁnd E (Y ). the number of times Brianna expects
to take the actuarial exam before she passes it. To determine E (Y) we’ll start with the formula l/(l — t) = l + t + t2 + l3 +  ‘  and go one step further. Taking the
derivative of both sides, we ﬁnd [as in Example 9.5(c)] that _2 _ 1 _d l g 2 3 H
(on t) (1) “ml dt[1_t] 1+2t+3t+4r+ . where this series likewise convergesT for it] < 1. Therefore, E(Y) : Z yPr(Y = y) = Z y(o.2)~‘“‘(0.8)
Fl i=1 2 (0.8) 2 y(0.2)y—l = (0.8)[1 + 2(02) + 3(0.2)2 + 4(0).)3 + A . ~]
yzl 1 ~ (0 8) l _ l _ 5
(l  0.2)2 ' (0.8)2 0.8 4
So Brianna expects to take the exam 1.25 times before she passes it. d) Finally, to determine Var(Y) we ﬁrst want to ﬁnd E ( Y2). To do so we ﬁrst multiply
the result in part (c) by t and ﬁnd [as in Example 9.5(c)] that t
(I —t)2 Differentiating both sides of this equation now gives us <1—r>2'<1>—t(2)(1«o<~1)_ 1+; d[ t ] — (0.8) — 1.25. :t+2t2+3t3+4t4+~. (1 —r)4 (I ~03 A d: (l —r)2 =12+22z+32r3+42z3+m. and this series is also convergenti for [II < 1. So now we have EM) : Z szr(Y = y) 2 Z yZ(0.2)«"*‘<0.8>
y=1 y=1 = (0.8) 2 ﬂow—1 = (0.8)[12 + 23(02) + 32(02)2 + 42(02)3 +  < .1
y=l _' 1+o.2 _ 1.2 _is
(amid—0.33] (0.8)3 8' 1LUsing the Ratio Test from calculus, one ﬁnds that . (n + 1)!"
11111 “900' ntn—l = mg}; " :1 2 m Ange (I + = M0) = ltl When I 2 i l, lim,Hoe nt"‘1 75 0, so the series does not converge for t = i 1. Consequently,
converges for M < 1. :0nce again we use the Ratio Test from calculus. Here this inﬁnite series (11 + 1 Ft” lim
n2tn—l "900 Z 2
2m lim ("+0 2m lim (1+3 =ltl(1)2=ltl. n—>oo "2 name When I = i l,lim,1_>00 rlzz"“1 75 0, so the series does not converge for t = i 1. Consequently, this inﬁnite series
converges for M < l. 430 Chapter 9 Generating Functions Consequently, to trial, after the ﬁrst, is independent of the outcome for a
bility of success for each Bernoulli trial is p, If we let the random variable Y count the n
then Y is a discrete random variable with pro ny previous trial. Further, the proba—
and the probability of failure is q = l — p.
umber of trials until we are ﬁnally successful,
bability distribution given by Pr(Y =y) :6]"T1p, y = 1, 2‘ 3, In addition, we ﬁnd that 1
E(Y) = — and Var(Y) = 31;.
p I p The following example uses the last identity in Table 9.2. (This identity was used earlier in
Examples 9.14 and 9‘ 15 ~ but rather implicitly.) EXAMPLE 9.19 Let f(x) = x/(l ~ x)“. This IS the generating function for the sequence a0, a1, a2, . . . , where ak = k for all k E N. The function g(x) = x(x + l)/(l — x)3 generates the sequence
b0, 191,173, .r .,forbk 2 k3, keN. The function h(x) = f (x)g(x) consequently gives us (10/90 + (610171 + a1b0)x + ((10192 + (21191 + 03170))62 +  ~ , so h(x) is the generating function for the sequence c0, c1,
62, . . . , where for each k E N, Ck =a0bk +albk~l +azbk—2 +"‘+ak~2b2 +Clk—ibi +akb0<
Here, for example, we ﬁnd that
00200220
c1 2012+102:0
c~2:022+l<12+203=l
c3=0<32+122+212+3~02=6 and, in general, ck :
Section Exercises.) Whenever a sequence c0, c1, Cg, . . . arises from two generating functions f (x) [for a0, a1, a2, . . .lanngc) [forb0, b1, b2, . . .], as in this example, the sequence c0, cl, c2, . ..
is called the convolution of the sequences a0, a1, azt . . . and b0, bi, b2, . . . . —‘\ Our last example provides one more instance of the convolution of sequences. EXAMPLE9.20 F3“ f(x):“(1‘)”:1+x+xz+x3+w and 8(x)=1/(1+x)=l—x+x2—
x +~,weﬁndthat f(X)g(x)=1/l(1—x)(1+x)l=1/(1—X2)=1+x2+x4+x6+~. {LO i (k — i)? (We shall simplify this summation formula in the 43] 9.2 Definition and Examples: Calculational Techniques Consequently, the sequence 1, 0, l, O, l, 0, , . , is the convolution of the sequences 1, l, l,
l, l, l,...andl,—l, 1,—1, l,~l,.... Mm 1. Find generating functions for the following sequences.
[For example, in the case of the sequence 0, l, 3, 9. 27, . . . ,
the answer required is x/(l — 3x), not 3i,r"+' or simply
0+x+3x2+9x3+~.l 2') (3% (it (3%” (3) 1” (fl 20% 30%  '  ~ 8(5) C) 1, Al, l, ~l, l, —l, . .. d) 0, 0,0, 6, —6, 6, —6, 6,...
e)1,0,1,0,1,0,1,,,. f) O,0,l,a,a2,a3,...,a7&0 2. Determine the sequence generated by each of the following
generating functions. 3) f(x) 2 (2X — 3)3
C) f(x) = x3/(l 7x3)
8) f(X):1/(3 —x)
f) f(x) = l/(l ﬁx) +3x7 ~ l1 3. In each of the following, the function f (x) is the generating
function for the sequence a0, a1, a3, . . , , whereas the sequence b) fix) =x4/(1— X)
d) f(x) : 1/(1+3x) 50, 191,192, . , . is generated by the function g(x). Express g(.r)
in terms of f(x).
a) b3 : 3
bn Zari,”€N,n b) 173 =
177 = 7
b" =an,nEN,n753,7
C) [91 2
[73 =
biz :2an,ﬂ€N,/’l 7+“ 1,3
d) bl 2
193 2
b7 :3 1),, =26l,1~l5,IZEN,117é 1,3,7
4. Determine the constant (that is, the coefﬁcient of XO) in
(3x2 '“ (2/x))15
5. a) Find the coefﬁcient of x7 in .
(1 +x +x2 +x3 + i  9‘5.
b) Find the coefﬁcient of x7 in
(1+x+x2+x3+~)”forn€Z+.
6. Find the coefﬁcient of x50 in» (x7 + x8 + x9 +  . 06.
7. Find the coefﬁcient ofx20 in (x2 + x3 + x4 + x5 + x6)5. 8. For It E Z+, ﬁnd in (l + x + x3)(l + x)” the coefﬁcient of
(a) 357; (b) x8; and (c) x’ forO 5 r f n + 2. r E Z. 9. Find the coefﬁcient ofx '5 in each of the following.
a) x3(l * 2X)“,
1)) (x3 — 5x)/(l — x)3
C) (1+XJ4/(l—XV
10. In how many ways can two dozen identical robots be as—
signed to four assembly lines with (a) at least three robots as signed to each line? (b) at least three, but no more than nine,
robots assigned to each line? 11. In how many ways can 3000 identical envelopes be divided,
in packages of 25, among four student groups so that each group
gets at least 150, but not more than 1000, of the envelopes? 12. Two cases of soft drinks, 24 bottles ofone type and 24 of an—
other, are distributed among ﬁve surveyors who are conducting
taste tests. [n how many ways can the 48 bottles be distributed
so that each surveyor gets (a) at least two bottles of each type?
(b) at least two bottles of one particular type and at least three
of the other? 13. If a fair die is rolled 12 times, what is the probability that
the sum of the rolls is 30'? 14. Carol is collecting money from her cousins to have a party
for her aunt. If eight of the cousins promise to give $2, $3, $4,
or $5 each, and two others each give $5 or $10, what is the
probability that Carol will collect exactly $40? 15. In how many ways can Traci select it marbles from a large
supply of blue, red, and yellow marbles (all of the same size) if
the selection must include an even number of blue ones? i 16. How can Mary split up l2 hamburgers and 16 hot dogs
among her sons Richard, Peter, Christopher, and James in such
a way that James gets at least one hamburger and three hot dogs,
and each of his brothers gets at least two hamburgers but at most
ﬁve hot dogs? 17. Verify that (l — x ~ x3 — x3 — x4 e x5 — x6)“ is the gen—
erating function for the number of ways the sum 11, where n e N,
can be obtained when a single die is rolled an arbitrary number
of times, 18. Show that ( l — 4x)‘V2 generates the sequence n E N. 19. a) If a computer generates a random composition of 8,
what is the probability the composition is a palindrome?
b) Answer the question in part (a) after replacing 8 by n, a
ﬁxed positive integer, 20. a) How many palindromes of 11 start with l? with 2‘? with
3? with 4? ...
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This note was uploaded on 05/12/2010 for the course MATH 2602 taught by Professor Costello during the Spring '09 term at Georgia Institute of Technology.
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