MATH 681
Problem Set #5
This problem set is due at the beginning of class on
November 10
.
1.
(5 points)
Find an asymptotically accurate approximation for
(
n
2
n
)
in terms of poly
nomials, exponentials, and selfexponentials. You may write it in bigO notation if you
wish.
By Stirling’s approximation,
±
n
2
n
²
≈
1
√
2
π
(
n
2
)
n
2
+1
/
2
(
n
2

n
)
n
2

n
+1
/
2
n
n
+1
/
2
We can simplify this a great geal with some simple algebra:
±
n
2
n
²
≈
1
√
2
π
n
2
n
2
+1
(
n
(
n

1))
n
2

n
+1
/
2
n
n
+1
/
2
≈
1
√
2
π
n
2
n
2
+1
(
n

1)
n
2

n
+1
/
2
n
n
2
+1
≈
1
√
2
π
n
(
n
2
)
(
n

1)
n
2

n
+1
/
2
≈
(
n

1)
n

1
/
2
√
2
π
±
n
n

1
²
(
n
2
)
The ﬁrst term in this product is clearly Θ(
n
n
√
n
). The second term is less obvious —
asymptotically it is a 1
∞
indeterminate form, and experimental veriﬁcation suggests it
grows very quickly. On the other hand, we know that
(
n
2
n
)
<
(
n
2
)
n
n
!
=
O
(
n
2
n
). Thus,
we can, with Stirling’s approximation, get the following (weak) asymptotic bounds:
Ω(
n
n
√
n
) =
±
n
2
n
²
=
O
(
n
2
n
)
2.
(10 points)
You have a large supply of beads of 4 diﬀerent colors and want to string
eight of them on a necklace, making use of each bead at least once. How many ways are
there to do so, if necklaces are considered identical if they are rotations or reﬂections
of each other?
The set
S
of necklaces prior to symmetryidentiﬁcation is the set of ordered choices of
8 items from a set of size 4, such that each item is chosen at least once. Then

S

is
the enumeration statistic 4!
S
(8
,
4) =
(
4
0
)
4
8

(
4
1
)
3
8
+
(
4
2
)
2
8

(
4
3
)
1
8
+
(
4
4
)
0
8
= 40824.
However, we seek

S/D
8

, where
D
8
is the dihedral group consisting of the identity, 7
rotations, and 8 reﬂections. We will invoke Burnside’s lemma to ﬁnd the invariants of
each of these permutations.
Every beadcoloring would remain ﬁxed under the identity, by deﬁnition, so its invari
ant set is exactly
S
itself, so it has 4!
S
(8
,
4) invariants.
Page 1 of 5
Due November 10, 2009
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View Full DocumentMATH 681
Problem Set #5
Under any of the odd rotations,
r
,
r
3
,
r
5
, or
r
7
, one may trace the positions of each
bead to note that the beads are mapped to eachother in an 8bead cycle, in which each
bead would have to have the same color as its successor, so every bead would have to
be the same color; there are no elements of
S
which match this condition, since every
coloring in
S
uses all 4 colors. Thus
r
,
r
3
,
r
5
, and
r
7
have no invariants (formally, one
might say they have 4!
S
(1
,
4) invariants, which happens to be zero).
Under the rotations
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 Fall '09
 WILDSTROM
 Approximation, Rotation, permute

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