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SIMON FRASER UNIVERSITY
DEPARTMENT OF MATHEMATICS
Midterm 2 – Solutions
MACM 201 Spring 2007 Instructor: Robert ˇS´amal March 7, 2007, 12:30 – 13:
[2] 1. Solve the following recurrence relation
an = 10an− 1 (n ≥ 1), a 2 = 5. We know that an = A · 10 n^ for some parameter A. By putting n = 2 we get a 2 = 5 = A · 102 , hence A = 1/ 20 , and
an = 20110 n^ (n ≥ 0).
[5] 2. (a) Solve the following recurrence relation. (More space is available on the next page.)
an+2 + 4an+1 + 4an = 0 (n ≥ 0), a 0 = 1, a 1 = 4.
The characteristic equation is t^2 + 4t + 4 = 0 and as t^2 + 4t + 4 = (t + 2)^2 , we have repeated root t = − 2. So, the general solution is of form an = A(−2)n^ + Bn(−2)n. We plug in n = 0, 1 :
1 = a 0 = A · 1 + B · 0 , hence A = 1 4 = a 1 = A · (−2) + B · (−2), hence B = − 3
It follows that an = (−2)n^ − 3 n(−2)n^ (n ≥ 0).
[2] (b) Check the answer for n = 2. Using the recurrence, we have a 2 = − 4 a 1 − 4 a 0 = − 4 · 4 − 4 · 1 = − 20. Using the formula we derived, we see a 2 = (−2)^2 − 3 · 2(−2)^2 = 4 − 24 = − 20.
[10] 3. For n ≥ 0 let an be the number of ways n can be expressed as a sum of odd integers,
if order matters. For example, a 4 = 3, as 4 = 3 + 1 = 1 + 3 = 1 + 1 + 1 + 1.
Find a recurrence relation for an (do not solve it). We group all expressions for n according to the last term.
- n = (... ) + 1 (the last term is 1). These expressions correspond precisely to expressions n − 1 = (... ), that is to different ways to write n − 1 as a sum of odd integers. There are an− 1 ways to do this.
- n = (... ) + k where k > 1 (the last term is not 1). These expressions precisely correspond to expressions n − 2 = (... ) + (k − 2), that is to different ways to write n − 2 as a sum of odd integers. There are an− 2 ways to do this.
So, we have an = an− 1 + an− 2 , (n ≥ 2). We still need the initial conditions: 1 = 1 is the only expression for 1, 2 = 1+1 the only expression for 2. Thus, a 1 = a 2 = 1.
For n = 0 it is not easy to make one’s mind if there is 0 or 1 way, but the recurrence relation gives a 0 = 0. From this we can come with a short solution (this was not asked for in the exam): an satisfies the same recurrence relation and the same initial conditions as the Fibonacci numbers, so an = Fn. Consequently,
an = √^1 5
)n −
)n^ ) (n ≥ 0).
5. Consider various walks in the graph on the figure. In parts (b), (c), (e) explain why the
walk you provided is not a path, a trail, and a cycle, respectively.
a
b
c
d
e
f
g
h
i
j
k
[2] (a) Write down an a-f path. Using various notations, we can write one such path as:
- a − d − e − f
- a, {a, d}, d, {d, e}, e, {e, f }, f
- a, ad, d, de, e, ef, f
- {a, d}, {d, e}, {e, f }
- ad, de, ef
In the next parts, we will only use the first notation.
[2] (b) Write down an a-f trail that is not a path. a − b − e − h − g − d − e − f (this is not a path, as the vertex e is used twice).
[2] (c) Write down an a-f walk that is not a trail. a − d − g − h − e − d − g − h − f (this is not a trail, as the edge {d, g} is used twice).
[2] (d) Write down an a-a cycle. a − b − e − d − a
[2] (e) Write down an a-a circuit (closed trail) that is not a cycle. a − b − e − h − f − e − d − a (this is not a cycle, as the vertex e is used twice).
