HW 6 Solutions
1.a
2x
= 2x(1+3x2 +(3x2 )2 +(3x2 )3 +. . .) = 0+2x+0x2 +6x3 +0x4 +18x5 +0x6 +54x7 +. . .
1 − 3x2
1.b
1
3
(1 − 3x) = 1 +
1
1
1
1
2
3
3
3
(−3x) +
(−3x) +
(−3x) + 3 (−3x)4 + . . . =
1
2
3
4
5
10
1 − x − x2 − x3 − x4 + . . .
3
3
3
1.c.
1
1
1
x
x
x
x
1
= ·
= (1 + + ( )2 + ( )3 + ( )4 + . . .) =
2−x
2 1 − 12 x
2
2
2
2
2
1 x x2
x3
x4
+ +
+
+
+ ...
2 4
8
16 32
1.d.
1
1
x4
log(1 + x2 )
2
= −log(
)·
=
−(−x
+
+. . .)(1+x+x2 +x3 +x4 +. . .) =
1−x
1 + x2 1 − x
2
1
1
0 + 0x + x2 + x3 + x4 + x5 + . . .
2
2
1.e.
1
1
1
1
1
(1+x2 +x3 ) 2 = (1+(x2 +x3 )) 2 = 1+ 2 (x2 +x3 )+ 2 (x2 +x3 )2 + 2 (x2 +x3 )3 +. . . =
1
2
3
1
1
1
1
1
1
1
1
1+ (x2 +x3 )− (x4 +2x5 +x6 )+ (x6 +3x7 +3x8 +x9 )+. . . = 1+ x2 + x3 − x4 − x5 − x6 +. . .
2
8
16
2
2
8
4
16
1.f. To solve this we compare the right and left sides of this equation, the left
side is
a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 + a7 x7 + a8 x8 + a9 x9 + . . . ,
while the right side expands to
1 + a0 x + 0x2 + a1 x3 + 0x4 + a2 x5 + 0x6 + a3 x7 + 0x8 + a4 x9 + . . . ,
setting the coefficients of each term equal gives
a0 = 1
a1 = a0 = 1
a2 = 0
a3 = a1 = 1
a4 = 0
a5 = a2 = 0
a6 = 0
a7 = a3 = 1
1
2
a8 = 0
a9 = a4 = 0
This gives that
A(x) = 1 + x + 0x2 + x3 + 0x4 + 0x5 + 0x6 + x7 + 0x8 + 0x9 + . . .
1.g.
(
∞
X
2
xn )3 = (1+x+x4 +x9 +. . .)3 = 1+3x+3x3 +x3 +3x4 +6x5 +3x6 +0x7 +3x8 +6x9 +. . .
n=0
The combinatorial interpretation of the mth coefficient of this generating function
is the number of weak composition of m into 3 squares.
1.h.
1
1
x2 x3 x4
x2 x3 x4
e + e−x
= (ex +e−x ) = [(1+x+ + + +. . .)+(1−x+ − + +. . .)] =
2
2
2
2! 3! 4!
2! 3! 4!
2
3
4
x
x
x
(1 + 0x +
+0 +
+ . . .)
2!
3!
4!
x
1.i.
ex
x2 x3 x4
x2
x3
x4
= (1+x+ + + +. . .)(1+x+x2 +x3 +x4 +. . .) = 1+2x+5 +16 +65 +. . .
1−x
2
6 24
2!
3!
4!
1.j.
x2
1
x2
1
x2
x3
x4
x5
x6
x2
+( )2 · +( )3 · +. . . = 1+0x+ +0 +3 +0 +15 +. . .
2
2
2!
2
3!
2!
3!
4!
5!
6!
The combinatorial interpretation of the mth coefficients is the number of permutations of n into cycles of length 2. This can be seen a result of the composition
2
formula for exponential generating functions since x2 is the exponential generating
function for the number of ways of putting n elements into exactly one 2-cycle (there
is one way if n=2, and zeros ways otherwise) and ex is the exponential generating
function for the number of ways of putting n two-cycles into permutation (there is
e
x2
2
= 1+
exactly one for all n). The composition of these two functions is e
the result.
x2
2
which gives
2.a. To make our computation easier we start by shifting the recurrence relation
we are trying to prove to:
S(n + 1, k + 1) = S(n, k) + (k + 1)S(n, k + 1).
We then consider the following calculations:
X
X
xn
xn
xn
∂ X
S(n, k) y k =
kS(n, k) y k−1 =
(k + 1)S(n, k + 1) y k
∂y
n!
n!
n!
n,k
n,k
n,k
and
∂ 1 X
xn
∂ X
xn
( (
S(n, k) y k ) − 1) =
S(n, k) y k−1 =
∂x y
n!
∂x
n!
n,k
n,k
3
X
n,k
S(n, k)
xn−1 k−1 X
xn
y
=
S(n + 1, k + 1) y k
(n − 1)!
n!
n,k
n
We then consider multiplying the reurrence relation by xn! y k , and summing over
n, k, using the above calculations and the generating function for S(n, k) gives that
the LHS simplifies to
x
∂ 1 y(ex −1)
(e
− 1) = ey(e −1) ex
∂x y
The RHS simplifies to
x
x
x
x
∂ y(ex −1)
ey(e −1) +
e
= ey(e −1) + ey(e −1) (ex − 1) = ey(e −1) ex
∂y
Thus the generating function for both sides of the relation are the same, so that
the relation holds.
2.b. The exponential generating function for n! is
X
X xn
1
=
xn =
.
n!
n!
1
−
x
n
n
By the formula for the product
of exponential generating functions, the exponential
P
generating function for k nk Dn−k is the product of the exponential generating
functions of Dn and the exponential
generating function for 1. The exponential
P n
generating function for 1 is n xn! = ex , and thus the product is
e−x
1
· ex =
.
1−x
1−x
Thus the exponential generating functions for both sides of the identity are equal,
and thus the identity holds.