Download Section 6.2 Calculating Coefficients Of Generating Functions

Section 6.2
Calculating Coefficients
Of Generating Functions
Aaron Desrochers
Ben Epstein
Colleen Raimondi
Tucker, Section 6.2
1
Calculating Coefficients Of
Generating Functions
• This chapter is about developing algebraic techniques
for calculating the coefficients of generating functions.
• All methods seek to reduce a given generating function
to a simple binomial –type generating function, or a
product of binomial-type generating functions.
Tucker, Section 6.2
2
Polynomial Expansions:
1)
1  x m1
 1  x  x 2  ...  x m
1 x
2)
1
 1  x  x 2  ...
1 x
3)
(1  x)n  1  C (n,1) x  C (n, 2) x 2  ...  C (n, r ) x r  ...  C (n, n) x n
4) (1  x m )n  1  C (n,1) x m  C (n, 2) x 2m  ...  (1) k C (n, k ) x km  ...  (1) n C(n, n) x nm
5)
1
2
r

1

C
(1

n

1,1)
x

C
(2

n

1,
2)
x

...

C
(
r

n

1,
r
)
x
 ...
n
(1  x)
6) If h(x)=f(x)g(x), where f(x)
h(x)
 a0  a1 x  a2 x 2  ...and
g(x)
 b0  b1 x  b2 x 2  ...,
then
 a0b0  (a1b0  a0b1 ) x  (a2b0  a1b1  a0b2 ) x 2  ...  (ar b0  ar 1b1  ar 2b2  ...  a0br ) x r  ...
Tucker, Section 6.2
3
6) If h(x)=f(x)g(x), where f(x)  a0  a1 x  a2 x 2  ... and g(x)  b0  b1 x  b2 x 2  ... , then
h(x)  a b  (a b  a b ) x  (a b  a b  a b ) x 2  ...  (a b  a b  a b  ...  a b ) x r  ...
0 0
1 0
0 1
2 0
1 1
0 2
r 0
r 1 1
r 2 2
0 r
• The rule for multiplication of generating functions in Eqn. (6) is
simply the standard formula for polynomial multiplication.
Tucker, Section 6.2
4
1)
1  x m1
 1  x  x 2  ...  x m
1 x
•Identity (1) can be verified by polynomial “long division”.
•We restate it, multiplying both sides of Eq.(1) by
As
(1  x)
(1  xm1 )  (1  x)(1  x  x 2  ...  x m )
We verify that the product of the right-hand side is
(1  xm1 )
by “long multiplication”
1  x  x 2  ...  x m
1 x
1  x  x 2  ...  x m
 x  x 2  x3 ...  x m  x m1
1
 x m 1
Tucker, Section 6.2
5
1)
1  x m1
 1  x  x 2  ...  x m
1 x
2)
1
 1  x  x 2  ...
1 x
• If m is made infinitely large, so that 1  x  x 2  ...  x m
becomes the infinite series 1  x  x 2  ...
then the
multiplication process will yield a power series in which the
coefficient of each x k , k  0 is zero.
We conclude that (1  x ) ( 1  x  x 2  ... )=1
[Numerically, this equation is valid for x  1 ;the
“remainder” term x m 1
Goes to zero as m becomes infinite.]
Dividing both sides of this equation by (1 – X) yields identity
(2).
Tucker, Section 6.2
6
3)
(1  x)n  1  C (n,1) x  C (n, 2) x 2  ...  C (n, r ) x r  ...  C (n, n) x n
Expansion (3), the binomial expansion was explained at the start
of section 6.1. Expansion (4) is obtained from (3) by expanding
(1  y)m , where y   x m
 n
 n
m
(1  ( x )]  1    ( x )    ( x m ) 2 
1 
 2
n
 n
m k
   ( x )     ( x m ) n
k
 n
m
n
Tucker, Section 6.2
7
5)
1
2
r

1

C
(1

n

1,1)
x

C
(2

n

1,
2)
x

...

C
(
r

n

1,
r
)
x
 ...
n
(1  x)
n
For identity 5, ( 1 – x)-n =
1
= ( 1 + x + x2 + … ) n
1-x
1
Since
1-x
= ( 1 + x + x2 + … )
(eq. 2)
2)
1
 1  x  x 2  ...
1 x
r
Let us determine the coefficient x in equation (7) by counting the number of
formal products whose sum of exponents is r, if ei represents the exponent of the
e
e
e
xe n
ith term in a formal product , the the number of formal products x 1 x 2 x 3
whose exponents sum to r is the same as the number of integer solutions to the
equation
e e e   e  r e  0
1
2
3
n
i
In example 5, section 5.4 we showed that the number of nonnegative integer
solutions to this equation is C(r + n –1,r ) , so the coefficient x r in eqn (7) is C(r
+ n –1,r ) . This verifies expansion (5).
Tucker, Section 6.2
8
1)
3)
5)
1  x m1
 1  x  x 2  ...  x m
1 x
(1  x)n  1  C (n,1) x  C (n, 2) x 2  ...  C (n, r ) x r  ...  C (n, n) x n
1
 1  C (1  n  1,1) x  C (2  n  1, 2) x 2  ...  C (r  n  1, r ) x r  ...
n
(1  x)
6) If h(x)=f(x)g(x), where f(x)  a
0
h(x)
 a1 x  a2 x 2  ...
and g(x)  b
0
 b1 x  b2 x 2  ...
, then
 a0b0  (a1b0  a0b1 ) x  (a2b0  a1b1  a0b2 ) x 2  ...  (ar b0  ar 1b1  ar 2b2  ...  a0br ) x r  ...
With formulas (1) and (6) we can determine the coefficients of a variety of
generating functions: first, perform algebraic manipulations to reduce a given
generating function to one of the forms (1  x)m ,(1  x m ) n , or (1  x)  n or
a product of two such expansions, then use expansions (3) and (5) and the
product rule (6) to obtain any desired coefficient.
Tucker, Section 6.2
9
Example 1
Find the coefficient of x in ( x  x  x 
16
2
3
4
)5
x16 in x10 (1  x)5 [i.e, the x6 termin (1  x)5 is
multiplied by x10 tobecomethe x16 termin x10 (1  x)5 ]
2
To simplify the expression, we extract x from each polynomial factor and the apply
identity (2).
( x 2  x3  x 4 
)5  [ x 2 (1  x  x 2 
 x10 (1  x  x 2 
1
10
x
(1  x)5
Thus the coefficient of
x16 in ( x 2  x3  x 4 
)]5
)5
)5is the coefficient of
x16 in x10 (1-x) –5 [i.e., the x6 term in (1-x) –5 is multiplied by to become
the x16 term in x10 (1-x) –5 ]
Tucker, Section 6.2
10
Example 1 continued
5)
1
2
r

1

C
(1

n

1,1)
x

C
(2

n

1,
2)
x

...

C
(
r

n

1,
r
)
x
 ...
n
(1  x)
From expansion (5) we see that the coefficient of
x6in (1  x)5 is C (6  5  1,6)
More generally, the coefficient of xr in
x r in x10 (1  x)5 equals the coefficient of x r 10 in
(1  x)5 , namely, C ((r  10)  5  1, (r  10)).
Tucker, Section 6.2
11
Example 2
6) If h(x)=f(x)g(x), where f(x)  a
0
h(x)
 a1 x  a2 x 2  ...
and g(x)  b
0
 b1 x  b2 x 2  ...
, then
 a0b0  (a1b0  a0b1 ) x  (a2b0  a1b1  a0b2 ) x 2  ...  (ar b0  ar 1b1  ar 2b2  ...  a0br ) x r  ...
Use generating functions to find the number of ways to collect $15 from 20
distinct people if each of the first 19 people can give a dollar (or nothing) and
the twentieth person can giver either $1 or $5 (or nothing).
The generating function for the number of ways to collect r dollars
from these people is (1+x)19(1+x+x 5). We want the coefficient of x15. The first
part of this generating function has the binomial expansion
(1+x)19 = 1 +
19
1
( )
x+
19
2
( )
x2+ … +
19 r
x+…+
r
( )
19
19
( )x
19
If we let f(x) be this first polynomial and let g(x) = 1+x+x5, then we can use Eq. (6)
to calculate the coefficient of x15 in h(x) = f(x)g(x). Let ar be the coefficient of xr in
f(x) in f(x) and br the coefficient of xr in g(x). We know that
ar =
19
r and that b0 = b1 = b5 = 1 (other bis are zero).
( )
Tucker, Section 6.2
12
Example 2 continued
6) If h(x)=f(x)g(x), where f(x)  a
0
h(x)
 a1 x  a2 x 2  ...
and g(x)  b
0
 b1 x  b2 x 2  ...
, then
 a0b0  (a1b0  a0b1 ) x  (a2b0  a1b1  a0b2 ) x 2  ...  (ar b0  ar 1b1  ar 2b2  ...  a0br ) x r  ...
Then the coefficient of of x15 in h(x) is
a15b0 + a14b1 + a13b2 + … + a0b15
Which reduces to
a15b0 + a14b1 + a10b5
Since b0, b1, b5 are the only nonzero coefficients in g(x). Substituting the
values of the various as and bs in Eq. (6), we have
19
15 x 1 +
( )
19
14
( )
x 1+
19
10 x 1 =
( )
19
15
19
14 +
19
10
( ) ( ) ( ).
+
Tucker, Section 6.2
13
Class Problem
How many ways are there to select 25 toys from seven types of toys with
between two and six of each type?
1  x m1
 1  x  x 2  ...  x m
1 x
1)
3)
4)
2)
1
 1  x  x 2  ...
1 x
(1  x)n  1  C (n,1) x  C (n, 2) x 2  ...  C (n, r ) x r  ...  C (n, n) x n
(1  xm )n  1  C (n,1) x m  C (n, 2) x 2m  ...  (1) k C (n, k ) x km  ...  (1) n C(n, n) x nm
5)
1
 1  C (1  n  1,1) x  C (2  n  1, 2) x 2  ...  C (r  n  1, r ) x r  ...
n
(1  x)
6) If h(x)=f(x)g(x), where f(x)  a
0
h(x)
 a1 x  a2 x 2  ...
and g(x)  b
0
 b1 x  b2 x 2  ...
, then
 a0b0  (a1b0  a0b1 ) x  (a2b0  a1b1  a0b2 ) x 2  ...  (ar b0  ar 1b1  ar 2b2  ...  a0br ) x r  ...
Tucker, Section 6.2
14
Class Problem (continued)
The generating function for ar, the number of ways to select r toys from
seven types with between 2 and 6 of each type, is
(x2 + x3 + x4 + x5 + x6)7
We want the coefficient of x25. We extract x2 from each factor to get
[x2 (1 + x + x2 + x3 + x4 )]7 = x14 (1 + x + x2 + x3 + x4 )7
Now reduce our problem to finding the coefficient of x 25-14 = x11 in
(1 + x + x2 + x3 + x4 )7. Using identity (1), we can rewrite this generating
function as
7
1 - x5 7
1
(1 + x + x2 + x3 + x4 )7 =
= (1 - x5)7
1-x
1-x
(
)
Tucker, Section 6.2
(
)
15
Class Problem (continued)
1  x m1
 1  x  x 2  ...  x m
1 x
1)
Let f(x) = (1 - x5)7 and g(x) = (1 - x5)-7 . By expansions (4) and (5), respectively,
we have
7 5
7 10
7 15
f(x) = (1 - x5)7 = 1 x +
x x +…
1
2
3
( )
g(x) =
( )
1
( )
( 1 - x ) = 1 + (1 + 17 – 1)x +(2 + 27 – 1)x
2
+…
+ r + 7 – 1 xr + …
r
(
)
Tucker, Section 6.2
16
Class Problem (continued)
To find the coefficient of x11, we need to consider only the terms in the product
1
of the two polynomials (1 - x5)7 and 1 - x that yield x11. The only nonzero
(
)
coefficients in f(x) = (1 - x5)7 with subscript < 11 (larger subscripts can be ignored)
are a0, a5, and a10. The products involving these three coefficients that yield x11
terms are:
a0b11
=1 x
(
+
11 + 7 – 1
+ 11
7
(
) ( 1))
a5b6
x
+
(
a10b1
6+7–1
+
6
7
1+7–1
x
(
)
(
).
) 2
1
Tucker, Section 6.2
17