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Sequences and the Difference Operator
Berton Earnshaw
February 16, 2005
1 Sequences
Number sequences appear all throughout mathematics. Examples are:
- 1 , 2 , 3 , 4 , 5 , 6 ,... (the natural numbers)
- 1 , 4 , 9 , 16 , 25 , 26 ,... (the square numbers)
- 3 , 9 , 27 , 81 , 243 , 729 ,... (the powers of 3)
- 1 , 2 , 3 , 5 , 8 , 13 ,... (the Fibonacci numbers)
- 1 , 2 , 4 , 7 , 12 , 20 ,... (shifted Fibonacci numbers)
- 1 , 2 , 8 , 42 , 262 , 1828 ,... (the meandric numbers)
Often we give the sequence a name or label so that we can easily reference it later. For instance, we may decide that an will represent the nth natural number. That is,
a 1 = 1, a 2 = 2, a 3 = 3, a 4 = 4, a 5 = 5, a 6 = 6,...
and in general an = n. Let bn represent the nth square number. Then
b 1 = 1, b 2 = 4, b 3 = 9, b 4 = 16, b 5 = 25, b 6 = 36,...
and in general bn = n^2. Let cn represent the nth power of 3. Then
c 1 = 3, c 2 = 9, c 3 = 27, c 4 = 81, c 5 = 243, c 6 = 729,...
and in general cn = 3n.
Notice that, using this formula for cn, it makes sense to talk about c 0 , that is, the 0th power of 3: c 0 = 3^0 = 1.
The same is true for an and bn:
a 0 = 0 and b 0 = 0^2 = 0.
These sequences are nice because we can write down a formula for the nth element in the sequence. That is, given the formula I can immediately calculate the nth element of the sequence. For instance, if I want to know the 277th square number, I simply compute
b 277 = 277^2 = 76729
or if I want to know the 13th power of 3, I compute
c 13 = 3^13 = 1594323
The Fibonnaci numbers are a little different. I begin by giving them a name, say fn for the nth Fibonacci number. Then
f 1 = 1, f 2 = 2, f 3 = 3, f 4 = 5, f 5 = 8, f 6 = 13,...
But how do I write down a formula for this sequence?!? Right now it is not obvious. However, it is obvious that
f 1 = 1, f 2 = 2, and fn = fn− 1 + fn− 2 for n > 2.
This is not a formula in the sense that I can immediately calculate the nth Fibonacci number from the sequence – the calculation requires knowledge of the previous two Fibonacci numbers. The equation fn = fn− 1 + fn− 2 is a called a recurrence relation for the Fibonacci numbers. In a sense, it is the next best thing to a formula – given the first couple of terms of the sequence, I can compute the rest rather quickly. One of the things we will explore in these lectures is the following: given a recurrence relation for a sequence, can we derive its formula? The answer is sometimes yes. In the case of Fibonacci numbers, it turns out that
fn =
)n+ −
)n+
2 n+
Call the nth shifted Fibonacci number gn. Then gn satisfies the recurrence relation g 1 = 1, g 2 = 2, and gn = gn− 1 + gn− 2 + 1 for n > 2
and the formula
gn =
)n −
)n
2 n+
1.1 Exercises
Write down a formula and/or a recurrence relation for the following sequences.
- 23, 23 , 23 , 23 , 23 , 23 ,...
1.2 Solutions
an = 23; a 1 = 23, an = an− 1 for n > 1
- 0, 1 , 8 , 27 , 64 , 125 , 216 ,... bn = n^3 ; b 0 = 0, bn+1 = bn + zn
- 7, 8 , 15 , 34 , 71 , 132 , 223 ,... cn = bn + 7 = n^3 + 7; c 0 = 7, cn+1 = cn + zn
- 0, 1 , 0 , − 1 , 0 , 1 , 0 , − 1 ,... xn = sin
n π 2
yn = n!
- 1, 7 , 19 , 37 , 61 , 91 ,... zn = bn+1 − bn = 3n^2 + 3n + 1; z 0 = 1, zn = zn− 1 + 6n for n > 1
Wow! The difference of cn is again zn. At first this may seem surprising, but actually it is quite obvious given the relationship between bn and cn. What are the differences of the other sequences in the exercises? Let an be the sequence in exercise 1. Then an = 23 and
∆an = an+1 − an = 23 − 23 = 0.
Let xn be the sequence in exercise 4. Then xn = sin(nπ/2) and
∆xn = xn+1 − xn
= sin
(n + 1) π 2
− sin
n π 2
= sin
n
π 2
π 2
− sin
n
π 2
= sin
n
π 2
cos
( (^) π 2
− cos
n
π 2
sin
( (^) π 2
− sin
n
π 2
= sin
n
π 2
· 0 − cos
n
π 2
· 1 − sin
n
π 2
= − cos
n π 2
− sin
n π 2
Using this formula to compute the first few elements of the difference sequence gives us
∆x 0 = − 1 , ∆x 1 = − 1 , ∆x 2 = 1, ∆x 3 = 1, ∆x 4 = − 1 , ∆x 5 = − 1 , ∆x 6 = 1, ∆x 7 = 1,...
which is exactly what we expect. Let yn represent the nth element in the sequence of exercise 5. Then yn = n!, and
∆yn = yn+1 − yn = (n + 1)! − n! = (n + 1) · n! − n! = (n + 1 − 1) · n! = n · n!
Finally, we know that zn = 3n^2 + 3n + 1. Thus
∆zn = zn+1 − zn = 3(n + 1)^2 + 3(n + 1) + 1 − (3n^2 + 3n + 1) = 3(n^2 + 2n + 1) + 3n + 3 + 1 − 3 n^2 − 3 n − 1 = 3n^2 + 6n + 3 + 3n + 3 + 1 − 3 n^2 − 3 n − 1 = 6n + 6 = 6(n + 1)
Isn’t this fun?!?
2.1 Exercises
Compute the difference of each of the following sequences.
- n^4
- en
- ln(n)
- n^3 ≡ (^) (nn−!3)! for n ≥ 3
(n 3
≡ (^) (n−n3)!3!! for n ≥ 3
3 More on the Difference Operator
This is fun, isn’t it? We now prove an important theorem about the difference operator.
Theorem 1. Let an and bn be sequences, and let c be any number. Then
- ∆(an + bn) = ∆an + ∆bn, and
- ∆(c · an) = c · ∆an.
Proof. We simply calculate
∆(an + bn) = (an+1 + bn+1) − (an + bn) = (an+1 − an) + (bn+1 − bn) = ∆an + ∆bn
and
∆(c · an) = c · an+1 − c · an = c · (an+1 − an) = c · ∆an
In the language of mathematics, Theorem 1 tells us that the difference op- erator is a linear operator. This is a very convenient property for an operator to have, as will become evident in the following proposition.
Proposition 1 (The difference of a polynomial). Let an be a polynomial in n of degree k ≥ 1. Then ∆an is a polynomial of degree k − 1.
We will begin by proving a short lemma.
Lemma 1. ∆nk^ =
∑k− 1 i=
(k i
ni^ for all k = 1, 2 , 3 ,.. ..
Proof. By the binomial theorem
(n + 1)k^ =
∑^ k
i=
k i
ni
k 0
k 1
n +
k 2
n^2 + · · · +
k k − 1
nk−^1 +
k k
nk.
Notice
(k k
= (^) (k−kk!)!k! = 1 so
∆nk^ = (n + 1)k^ − nk
∑^ k
i=
k i
ni^ − nk
k∑− 1
i=
k i
ni^ +
k k
nk^ − nk
k∑− 1
i=
k i
ni
Proof of Proposition 1. If an is a polynomial in n of degree k, then
an =
∑^ k
i=
cini^ = c 0 + c 1 n + c 2 n^2 + · · · + ck− 1 nk−^1 + cknk
for some constants c 0 , c 1 ,... , ck where ck 6 = 0. Then by the linearity of the difference opeartor
∆an = ∆
∑^ k
i=
cini^ =
∑^ k
i=
ci∆ni.
We just saw in Lemma 1 that ∆ni^ is a polynomial of degree i − 1 for i ≥ 1. For i = 0, ∆n^0 = ∆1 = 0. Therefore, ∆an is the sum of polynomials of degrees 0 through k − 1. Thus ∆an is a polynomial of degree k − 1.
Now it is your turn to compute the difference of a few general sequences. Use the linearity of the difference operator when possible.
3.2 Solutions
∆c · bn^ = c∆bn = c(bn+1^ − bn) = c(b · bn^ − bn) = c(b − 1)bn
∆c logb(n) = c∆ logb(n) = c (logb(n + 1) − logb(n))
= c logb
n + 1 n
= c logb
n
∆nk^ = (n + 1)k^ − nk
=
(n + 1)! (n + 1 − k)!
n! (n − k)!
=
(n + 1)! (n + 1 − k)!
(n + 1 − k) · n! (n + 1 − k)!
= (n + 1) · n! − (n + 1 − k) · n! (n + 1 − k)!
=
(n + 1 − (n + 1 − k)) · n! (n + 1 − k)!
=
k · n! (n + 1 − k)!
= k
n! (n − (k − 1))! = knk−^1
- This follows easily from the previous exercise since
(n k
= (^) (n−nk!)!k! = (^) k^1! nk:
n k
k!
nk
k!
∆nk
= k k!
nk−^1
(k − 1)!
n! (n − (k − 1))!
=
n! (n − (k − 1))!(k − 1)!
=
n k − 1
One could also see this by considering Pascal’s triangle.
- Give the formula for a sequence cn such that ∆cn = n^3. (hint: write n^3 as a linear combination of n^3 , n^2 and n^1 )
- Give the formula for a sequence dn such that ∆dn = dn. (hint: consider an exponential sequence)
3.4 Solutions to Take-Home Exercises
- an = 12 n^2 = n(n 2 − 1), for example
- bn = 13 n^3 + 12 n^2 = n(n−1)(2 6 n−1), for example
- cn = 14 n^4 + n^3 + 12 n^2 = n
(^2) (n−1) 2 4 , for example
- dn = 2n, for example
