MA 3260 Practice Final, Part I Answers in Red
Name
December 2, 2014
Note: Final is scheduled for Thursday, December 12th at 10:45am.
Recall that the recurrence relation for the Fibonacci sequence is
1.
(1)
What is the characteristic equation for the Fibonacci sequence? x2 − x − 1 = 0
a.
b.
c.
An − An−1 − An−2 = 0.
Find the two basic solutions.
√ n
√ n
An = 1+2 5
and An = 1−2 5
Find the general solution.
√ n
√ n
An = B · 1+2 5
+ C · 1−2 5
d.
The standard Fibonacci sequence starts off with the terms A0 = 1 and A1 = 1. Find the particular
solution, if instead, the first two terms are A0 = 1 and A1 = 0.
I got B =
√
5−1
√
2 5
(2)
and C =
√
5+1
√ ,
2 5
which gives us a particular solution of
√
√ !n √
√ !n
5−1
1+ 5
5+1
1− 5
√ ·
An =
+ √ ·
.
2
2
2 5
2 5
For problems 2-4, find the basic solution, the general solution, and the particular solution.
2.
The sequence 2, 3, 3, −3, −33, −147, . . . with recurrence relation An − 5An−1 + 6An−2 = 0. Particular
solution: An = 3 · 2n − 3n .
3.
The sequence 1, 0, −4, 16, −48, . . . with recurrence relation An +4An−1 +4An−2 = 0. Particular solution:
An = (−2)n − n(−2)n .
√
√
√
4.
The sequence 1, 2 − 2 3, −8, −8 − 8 3,√
. . . with recurrence relation An − 2 3An−1 + 4An−2 = 0.
Particular solution: An = 2n cos( nπ
) − (2 − 3 3) · 2n sin( nπ
).
6
6
5.
Write out the first 11 rows of Pascal’s triangle, and expand the following.
a.
(x + 1)8 .
x8 + 8x7 + 28x6 + 56x5 + 70x4 + 56x3 + 28x2 + 8x + 1
b.
(a − b)5 .
a5 − 5a4 b + 10a3 b2 − 10a2 b3 + 5ab4 − b5
c.
(a + 2b)5 .
a5 + 5a4 2b + 10a3 22 b2 + 10a2 23 b3 + 5a24 b4 + 25 b5 = a5 + 10a4 b + 40a3 b2 + 80a2 b3 + 80ab4 + 32b5
1
MA 3260 Practice Final, Part I Answers in Red
6.
2
Find an optimal Eulerization for the following graphs.
We need at least 5 edges to fix 10 odd vertices, so this Eulerization must be optimal.
We need at least 3 edges to fix 6 odd vertices, but the two lower odd vertices aren’t adjacent to another odd,
so each will need an additional edge. Therefore, 5 edges is best possible.
MA 3260 Practice Final, Part I Answers in Red
7.
a.
3
Consider the graph consisting of the four sides and four corners of a 2 by 8 rectangle.
Find a minimal spanning tree for this graph. What is the total length of the minimal spanning tree.
Take the two sides of length 2, and one of the sides of length 8, for a total length of 8 + 2 + 2 = 12.
b.
Now add two Steiner points along the mid-line of the rectangle. What is the length of the resulting
Steiner tree. (If the coordinates of the rectangle are (0, 0), (8, 0), (8, 2), and (0, 2), then the two Steiner points
would have y coordinate equal to 1.) Your two Steiner points will have coordinates ( √13 , 1) and (8 − √13 , 1).
The five edges will consist of four with length √23 and one of length 8 − 2 · √13 . The total length is 11.464 . . ..
8.
Given a graph that has a minimal spanning tree of total length 100, and a Steiner tree of length 98,
what do you know about the length of the shortest possible network joining the vertices of the graph? (Note
there may be more than one Steiner tree)
•
•
•
•
•
It’s definitely 100.
It’s definitely 98.
It’s definitely shorter than 98.
It’s either 98 or shorter. There could be another Steiner network that is shorter.
It could be greater than 100.
9.
In Problem 6, suppose that you know that there is only one possible Steiner tree. What’s your answer
now?
• It’s definitely 100.
• It’s definitely 98. The shortest network is either a Steiner tree or a minimal spanning tree. The
information given tells you that the shortest network is not the minimal spanning tree.
• It’s definitely shorter than 98.
• It’s either 98 or shorter.
• It could be greater than 100.
10.
How many edges does a Steiner point have?
•
•
•
•
•
11.
Two
Three A Steiner point always has three edges.
Four
Could be any number less than four.
Could be any number.
What possible angles can the edges of a Steiner point make?
•
•
•
•
•
30◦ , 45◦ , 60◦ , or 90◦ .
Only 120◦’s. A Steiner point always has three edges that make three 120◦ angles.
Could be any angle.
Either three 120◦’s or four 90◦’s.
Only 90◦ ’s.