Download Exponential Generating Functions

Exponential Generating Functions
Book Problems
22. Determine the exponential generating function for the sequence of factorials 0!, 1!, 2!, . . . , n!, . . .
∞
X
k=0
xk =
∞
X
k!
k=0
1
xk
=
k!
1−x
28. Determine the number of n-digit numbers with all digits at least 4, such that 4 and 6 occur
an even number of times, and 5 and 7 each occur at least once, there being no restrictions on
the digits 8 and 9.
2 2 2
x2 x4 x6
x2 x3
x2 x3
1+
+
+
+ ...
x+
+
+ ...
1+x+
+
+ ...
2!
4!
6!
2!
3!
2!
3!
In closed form, this is
1 x
e + e−x
2
The answer we seek will be the coefficient of
2
(ex − 1)2 e2x
xn
n! .
Additional Problems
1. For each of the following sequences {an }, find a simple, closed form expression for the exponential generating function.
(a) (5, 5, 5, . . .)
5ex
(b) ak = 3k
e3x
(c) (1, 0, 0, 1, 1, . . .)
ex − x −
x2
2!
(d) (0, 0, 1, 1, . . .)
ex − 1 − x
(e) (1, 0, 1, 0, 1, . . .)
1 x
e + e−x
2
(f) (2, 1, 2, 1, 2, 1, . . .)
1 x
3
1
ex + e−x +
e − e−x = ex + e−x
2
2
2
2. How many 10-letter words are there in which each letter e,n,r and s occur
(a) At most one?
22
x2
1+x+
+ ...
(1 + x)4
2!
We need the coefficient of
x10
10! .
To find this, we will use the closed form and some algebra.
(1 + 4x + 6x2 + 4x3 + x4 )e22x = e22x + 4xe22x + 6x2 e22x + 4x3 e22x + x4 e22x
We have to factor in that there is a shift, so 10! will not be associated with x10 . We have
10
to use the ‘multiplication’ trick to get x10! .
• For e22x , the coefficient is 2210
10
• For 4xe22x , we have x9! , so we need to multiply by 10
10 to get the right term. So, the
coefficient would therefore be 4 · 10 · 229 = 40 · 229 .
10
9·10
• For 6x2 e22x , we have x8! , so we need to multiply by 9·10
to get the right term. So,
8
the coefficient would therefore be 6 · 9 · 10 · 22 = 540 · 228 .
10
• For 4x3 e22x , we have x7! , so we need to multiply by 8·9·10
8·9·10 to get the right term. So,
7
the coefficient would therefore be 4 · 8 · 9 · 10 · 22 = 2880 · 227 .
10
• For x4 e22x , we have x6! . So, we need to multiply by 7·8·9·10
7·8·9·10 , so the coefficient would
6
6
therefore be 7 · 8 · 9 · 10 · 22 = 5040 · 22 .
So, the coefficient of
x10
10!
will be
2210 + 40 · 229 + 540 · 228 + 2880 · 227 + 5040 · 226
(b) At least once?
22 4
x2
x2 x3
1+x+
+ ...
x+
+
+ ...
2!
2!
3!
We need the coefficient of
x10
10! .
We again will use closed forms and algebra.
e22x (ex − 1)4 = e22x (e4x − 4e3x + 6e2x − 4ex + 1)
= e26x − 4e25x + 6e24x − 4e23x + e22x
So, the coefficient of
x10
10!
will be
2610 − 4 · 2510 + 6 · 2410 − 4 · 2310 + 2210