18.314 PSET 3 ANSWER KEY, GRADING CRITERIA
November 17, 2007
General Ideas For Grading
• -1 pt. for calculation error.
• -2 ∼ -4 pts. if only one step further to the answer.
• -6 ∼ -8 pts. if get the very basic idea but no constructive progress.
• 0 pt.!!!!!! if misunderstand the whole problem.
Hints and Answers
Problem 1. (a) F (x) =
an = 3n − 2n
x
6x2 −5x+1
(b) G(x) = e3x − e2x (c)
Hints:
(a) Multiply the original by xn+2 and sum over all n ≥ 0, we get the relation in
term of F(x).
n
(b) Multiply the original by xn! and sum over all n ≥ 0.
(c) Use part (b) and consider the Taylor expansion of the function, then we can
get the coefficient of xn which supposed to be an!n .
Problem 2. (a) F (x) = (2x + 7)ex − 6e2x + 5e3x (b)
bn = 2n + 7 − 6 · 2n + 5 · 3n
Hints:
(a) By multiplying the recursion relation by
derive the differentiation equation:
xn
n!
and take the summation, we can
F 00 (x) = 5F 0 (x) − 6F (x) + 4(x + 2)ex
(b) Consider the Taylor expansion for eax and use the fact that
F (x) =
∞
X
bn xn
n=0
1
n!
Problem 3. A = 7 , B = −12
Hints: Either one of the following ways works:
• Randomly put two different n and solve for a linear equation system with
two variable.
• Use the fact that 3 and 4 should be the roots of x2 − Ax + B = 0.
Problem 4. F (x) =
x9
2
(1−x) (1−x2 )2 (1−x3 )2
Hints: Let’s say t(n) is the way to partition n into at most three part, so that
one can express n = 3a1 + 2a2 + a3 , a1 , a2 , a3 ≥ 0 and theP
three parts of n are
a1 + a2 + a3 ≥ a1 + a2 ≥ a1 . Therefore, if we say G(x) = t(n)xn , then
1
G(x) = (1 + x + x2 + · · · )(1 + x2 + x4 + · · · )(1 + x3 + x6 + · · · ) = (1−x)(1−x
2 )(1−x3 ) .
Now, we have
n−9
X
p(n, 3) =
t(i)t(n − 9 − i)
i=0
, by some further simplification, we would have F (x) = x9 · (G(x))2 .
2
Problem 5. F (x) = e
x
(x+ 2(1−x)
)
Hints: Denoted
n! gn as the number of paths on a k vertices graphs. Once can see
2 , n ≥ 2; Therefore, the exponential generating function of g is
that gn =
n
1, n = 1.
2
3
G(x) = x + x2 + x2 + · · · .
Now, since we can partition the graph into many paths so F(x) should be the
x2
exponential formula on G(x). F (x) = eG(x) = e(x+ 2(1−x) ) .
Problem 6. exe
x
Hints: Consider removing the vertex 1, so every component of the rest graph
would have depth ≤ 1. Therefore, one can find out that exponential generating
x
xn−1
function of the component is xex and the gn is the coefficient of (n−1)!
in exe .
Problem 7.
Hints: Number the vertices of (n+2)-gon form 1 through n+2 and let 2k be the
first x which the path touch the x-axis then (1,2,k+2) will be the first triangle
we cut off from the (n+2)-gon. Now, the way to triangulate the (k+1)-gon with
vertex 2 to vertex k+2 is using the path from (1,1) to (2k-1,1) and (n-2k+2)-gon
2
with vertices k+2 through 1. Therefore, by induction, this is a bijection.
Problem 8.
Hints: Either using combinatorics way to prove that the number of Dyck paths
of length 2n that start with exactly k up steps and the number of Dyck paths of
length 2n
have
that
exactly k+1 points on the horizontal x-axis are both
2n−k−1
2n−k−1
−
or use bijection to prove that the number of Dyck paths of
n−1
n
length 2n that start with at least k up steps is equal to the number of Dyck
paths of length 2n that have at least k+1 points on the horizontal x-axis.
(PS: the roughly description of the bijection is as follows: if a path that touches
the x-axis at least k+1 times, then remove the up steps at the first k-1 places it
touches the x-axis without the origin, and place these up steps at the beginning
of the path.)
Problem 9*.
Hints: Show buy each time adding a number k to every number of the set that
the number of vectors is C(2n,n)
n+1 .
Problem 10*. Cn−1 · (n − 1)!
Hints:
Consider F(x) be the exponential generating function for the
P
weight(T) and try to get the relation that F 0 (x) = 1+xF1 0 (x)
Contact: shinnyih@mit.edu
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