4 Sequences as Functions
We’ve seen many different types of function in this course and used them to model various situations.
In practice, one is often faced with the opposite problem: given experimental data, what type of
function should you try?
4.1 Polynomial Sequences: First, Second, and Higher Differences
To begin to answer this, first ask yourself, “What is a sequence?” Hopefully you have a decent
intuitive idea already. More formally, a sequence a function whose domain is a set like the natural
numbers, for example
f : N → R : n 7→ 3n
2
−2
defines the sequence
f (1), f (2), f (3), . . .
=
1, 10, 25, 46, 73, . . .
This is indeed the intuitive idea of a function to many grade-
school students: continuity and domains including fractions
or even irrational numbers are more advanced concepts.
Suppose instead that all you have is a data set
x 1 2 3 4 5
y 1 10 25 46 73
0
20
40
60
80
y
0 1 2 3 4 5
x
perhaps arising from an experiment. Could you recover the original function y = f (x) directly from
this data? You could try plotting data points as we’ve done, though it is hard to decide directly
from the plot whether we should try a quadratic model, some other power function/polynomial, or
perhaps an exponential. Of course, the physical source of real-world data might also provide clues.
A more mathematical approach involves considering how data values change:
x 1
+1
''
2
+1
''
3
+1
''
4
+1
''
5
y 1
+9
55
10
+15
55
25
+21
55
46
+27
55
73
+6
33
+6
33
+6
33
The first-differences in the x-values are constant whereas those for the y-values are increasing
y
n+1
− y
n
=
9, 15, 21, 27, . . .
You likely already notice the pattern: the sequence of first-differences is increasing linearly as the
arithmetic sequence
y
n+1
− y
n
= 3 + 6n
To make this even clearer, note that the sequence of second-differences in the y-values is constant
(+6). These facts are huge clues that we expect a quadratic function.
44
But why? Well we can certainly check the following directly:
Linear Model If f (n) = an + b, then the sequence of first-differences is constant
f ( n + 1) − f (n) = a
Quadratic Model If f (n) = an
2
+ bn + c, then the sequence of first-differences is linear and the
second-differences are constant:
g(n) := f (n + 1) − f (n) = 2an + a + b, g(n + 1) − g(n) = 2a
The relationship between these results and the derivative(s) of the original function f (x) should feel
intuitive: what happens if you differentiate a quadratic twice?
Example 4.1. You are given the following data set
x 0 2 4 6 8 10
y 2 16 22 20 10 −8
The x-values have constant first-differences while the y-values
have constant second-differences
First-differences: 14, 6, −2, −10, −18
Second-differences: −8, −8, −8, −8
0
10
20
y
2 4 6 8 10
x
We therefore suspect a quadratic model y = f (n) = an
2
+ bn + c. Rather than using the above
formulae, particularly since the x-differences are not 1, it is easier just to substitute:
2 = y(0) = c,
(
16 = f (2) = 4a + 2b + 2
22 = f (4) = 16a + 4b + 2
=⇒
(
2a + b = 7
8a + 2b = 10
=⇒ 4a = −4
whence a = −1, b = 9 and c = 2. A quadratic model is therefore
y = f (n) = −n
2
+ 9n + c = −n
2
+ 9n + 2
It is easily verified that the remaining data values satisfy this relationship.
There are at least two issues with our method:
1. The question we’re answering is, “Find a quadratic model satisfying given data.” Constant
second-differences don’t guarantee that only a quadratic model is suitable. For example,
y = −n
2
+ 9n + 2 + 297n(n −2)(n − 4)(n − 6)(n − 8)(n −10)
is a very complicated model satisfying the same data set!
2. It is very unlikely that experimental data will fit such precise patterns (why not?). However,
if the differences are close to satisfying such patterns, then you should feel confident that a
linear/quadratic model is a good choice.
45
Example 4.2. Given the data set
x 0 2 4 6 8 10
y 3 23 41 59 77 93
with sequences of first- and second-differences
First-differences: 20, 18, 18, 18, 16
Second-differences: −2, 0, 0, −2
0
20
40
60
80
y
0 2 4 6 8 10
x
do you think a linear or quadratic model would be superior?
If you wanted a linear model, you’d likely be inclined to try f (x) = 9x + b for some constant b. Here
are two options:
1. f (x) = 9x + 5 fits the middle four data values perfectly, but as a predictor is too large at the
endpoints: f (0) = 5 > 3 and f (10) = 95 > 93.
2. f (x) = 9x + 5 −
2
3
doesn’t pass through any of the data values but seems to reduce the net error
to zero:
x 0 2 4 6 8 10
f (x) − y −
4
3
2
3
2
3
2
3
2
3
−
4
3
=⇒
∑
x
f (x) − y = 0
Neither model is perfect, but then this is what you expect with real-world data!
Exercises 4.1. 1. For each data set, find a function y = f (x) modelling the data.
(a) x 2 4 6 8
y −1 2 7 14
(b) x 2 5 8 11 14
y −6 −15 −6 21 66
(c) x 0 6 9 15
y 3 15 21 33
(Be careful with (c): the x-differences aren’t constant!)
2. Suppose a table of data values containing (x
0
, y
0
) has constant first-differences in both variables
∆x = x
n+1
− x
n
= a, ∆y = b
Find the equation of the linear function y = f (x) through the data.
3. What relationship do you expect to find with the sequential differences of a cubic function
f ( n) = an
3
+ bn
2
+ cn + d? What about a degree-m polynomial f (n) = an
m
+ bn
m−1
+ ···?
4. If f (n) = an
2
+ bn + c is a quadratic model for the data in Example 4.2 with constant second-
differences −1, show that a = −
1
8
. What might be reasonable values for b, c?
5. (Hard) Suppose f (x) is a twice-differentiable function and h > 0 is constant. Use the mean
value theorem from calculus to explain the following.
(a) First-differences f (x + h) − f (x) are proportional to f
′
(ξ) for some ξ ∈ (x, x + h).
(b) Second-differences satisfy
f (x + 2h) − f (x + h)
−
f (x + h) − f (x)
= f
′′
(ξ)hα for some
ξ between x and x + h and some α. Why is it unlikely that α is constant?
46
4.2 Exponential, Logarithmic & Power Sequences
To observe relationships between data values, you might also have to consider ratios between succes-
sive terms or skip values.
Example 4.3. From a first glance at the given data, it is hard to decide whether an exponential or
a quadratic (or higher degree polynomial) model is more suitable. If we try to apply the constant-
difference method, we don’t seem to get anything helpful:
x 1
+2
((
3
+2
((
5
+2
))
7
y 15
+120
33
135
+1080
22
1215
+9720
11
10935
+960
11
+8640
22
By the time we’re looking at second-differences, any conclusion
would be very weak since we only have two data values!
0
2
4
6
8
10
y (1000s)
0 2 4 6
x
If instead we think about ratios of y-values, then a different pattern emerges:
x 1
+2
((
3
+2
((
5
+2
))
7
y 15
×9
33
135
×9
22
1215
×9
11
10935
The question remains: what type of function scales its output by 9 when 2 is added to its input:
f (x + 2) = 9 f (x)? This is a function that converts addition to multiplication: an exponential! If we try
y = f (x) = ba
x
for some constants a, b, then
f (x + 2) = ba
x+2
= ba
2
b
x
= a
2
f (x)
from which a suitable model is y = 5 · 3
x
.
We can see the pattern in the example more generally:
Exponential Model If f (x) = ba
x
, then adding a constant to x results in
f (x + k) = ba
x+k
= a
k
f (x)
If x-values have constant differences (+k), then y-values will be related by a constant ratio (×a
k
).
You might remember this as ‘addition–product’ or ‘arithmetic–geometric.’
Such a simple pattern is often disguised:
• Complete data might not be given so you might have to skip some data values to see a pattern.
For example, if our original data was
x 1 3 4 5 7
y 15 135 405 1215 10935
then the x-values are not in a strictly arithmetic sequence.
• As in Example 4.2, real-world/experimental data will only approximately exhibit such patterns.
47
Example 4.4. A population of rabbits is measured every two months resulting in the data set
t 0 2 4 6 8 10
P 5 7 10 14 19 28
The data seems very close to being quadratic; consider the first and second sequences of P-differences
∆P =
2, 3, 4, 5, 9
, ∆∆P =
1, 1, 1, 4
However, the last difference doesn’t fit the pattern. Instead, the fact that we expect an exponential
model is buried in the experiment: the data is measuring population growth! We therefore instead
consider the ratios of P-values:
t 0
+2
''
2
+2
''
4
+2
''
6
+2
''
8
+2
))
10
P 5
×1.4
77
7
×1.43
55
10
×1.4
55
14
×1.36
55
19
×1.47
55
28
The ratios are very close to being constant, whence an exponential model is suggested! To exactly
match the first and last data values, we could take the model
P(t) ≈ 5
28
5
t
10
t 0 2 4 6 8 10
P 5 7.057 9.960 14.057 19.839 28
Only P(8) doesn’t match when we take rounding to the nearest integer into account.
We’ve seen that addition-addition corresponds to a linear model and that addition-multiplication to
an exponential. There are two other natural combinations.
Logarithms These operate exactly as exponentials but in reverse. If f (n) = log
a
x + b, then multiply-
ing x by a constant results in a constant addition/subtraction to y:
f ( kx) = log
a
(kx) + b = log
a
k + log
a
x + b = log
a
k + f (x)
This could be summarized as ‘product–addition.’
Power Functions If f (x) = ax
m
, then multiplying x by a constant will do the same to y
f ( kx) = a(kx)
m
= ak
m
x
m
= k
m
f (x)
We have a ‘product–product’ relationship between successive terms.
Examples 4.5. Find the patterns in the following data and suggest a model y = f (x) in each case.
x 6 18 54 162
y 1 2 3 4
x 3 6 9 12
y 135 1080 3645 8640
The sequential approach in this chapter is a form of discrete calculus: using a pattern of differences to
predict the original function is similar to how we use knowledge of a derivative f
′
(x) to find f (x).
48
Example 4.6. Suppose g(2) = 3 and g(4) = 9. What do you think should be the value of g(8)?
It depends on the type of model you try.
1. For a linear (addition-addition) model we know that
∆x = 2 corresponds to ∆y = 6, so
g(8) = g(4 + 2∆x) = g(4) + 2∆y = 9 + 12 = 21
2. For an exponential (addition-product) model, ∆x = 2
corresponds to a y-ratio r
y
=
9
3
= 3, so
g(8) = r
y
g(6) = r
2
y
g(4) = 9 ·9 = 81
3. For a power (product-product) model, r
x
= 2 corre-
sponds to a r
y
= 3, so
g(8) = g(2 ·4) = g(4r
x
) = r
y
g(4) = 3 ·9 = 27
0
20
40
60
80
y
0 2 4 6 8
x
1. g(x) = 3x − 3
2. g(x) = 3
x/2
3. g(x) = x
log
2
3
We do not need to calculate the models explicitly(!), though they are stated below the graph for
convenience.
Exercises 4.2. 1. Find the patterns in the following data sets and use them to find a model y = f (x).
(a) x 0 1 2 3 4
y 80 120 180 270 405
(b) x 2 4 8 10
y 1 16 256 625
(c) x 1 3 5 7 9
y 15 5 19 57 119
(d) x 1 3 4 6
y 1 36 216 7776
(e) x 20 60 180 540
y 2 4 6 8
(f) x 2 6 54 486 4374
y 2 4 8 12 16
2. Take logarithms of the power relationship y = ax
m
. What is the relationship between ln y and
ln x? Use this to give another reason why the inputs and outputs of power functions satisfy a
‘product–product’ relationship.
3. How does our analysis of exponential functions change if we add a constant to the model? That
is, how might you recognize a sequence arising from a function f (x) = ba
x
+ c?
4. Suppose f (5) = 12 and f (10) = 18. Find the value of f (20) supposing f (x) is a:
(a) Linear function;
(b) Exponential function;
(c) Power function.
If f (20) = 39, which of the three models do you think would be more appropriate?
49
4.3 Newton’s Method
To finish our discussion of sequences we revisit a (hopefully) familiar technique for approximating
solutions to equations. Variations of this approach have been in use for thousands of years.
Example 4.7. We motivate the method by considering an ancient method for approximating
√
2,
known to the Babylonians 2500 years ago!
Suppose x
n
>
√
2. Then
2
x
n
<
2
√
2
=
√
2. It seems reasonable to guess that their average
x
n+1
=
1
2
x
n
+
2
x
n
should be a more accurate approximation to
√
2. If start with an initial guess x
0
= 2, then we obtain
the sequence
x
1
=
1
2
2 +
1
2
=
3
2
, x
2
=
17
12
= 1.4166 . . . , x
3
=
577
408
= 1.4142 . . . , . . .
This sequence certainly appears to be converging to
√
2. . .
Since it makes use of the average, this approach is sometimes called the method of the mean. It may be
applied to any square-root
√
a where a > 0: let x
0
> 0 and define,
x
n+1
:=
1
2
x
n
+
a
x
n
(∗)
A rigorous proof that the sequence converges requires more detail than is appropriate for us (though
see Exercise 3), but two observations should make it seem more believable:
1. If the sequence (∗) has a limit L, then the limit must satisfy
L =
1
2
L +
a
L
=⇒ 2L
2
= L
2
+ a =⇒ L
2
= a =⇒ L =
√
a
where we take the positive root since all terms x
n
are plainly positive.
2. The iterations have a convincing geometric interpretation.
The sequence of iterates can be found by repeatedly tak-
ing the tangent line to the curve y = f (x) = x
2
− a and
intersecting it with the x-axis. To see why, observe that
the tangent line at x
n
has equation
y = f (x
n
) + f
′
(x
n
)(x − x
n
)
= x
2
n
− a + 2x
n
(x − x
n
)
= 2x
n
x − x
2
n
− a
which intersects the x-axis (y = 0) when
x =
x
2
n
+ a
2x
n
=
1
2
x
n
+
a
x
n
= x
n+1
0
y = x
2
− a
−a
x
n
x
n+1
50
This geometric idea generalizes. . .
Definition 4.8. Given a differentiable function f (x) with non-zero derivative, the Newton–Raphson
iterates of an initial value x
0
are defined by the recurrence formula
x
n+1
:= x
n
−
f (x
n
)
f
′
(x
n
)
Our two previous observations still hold:
1. If L = lim
n→∞
x
n
exists and f
′
(L) = 0, then
L = L −
f (L)
f
′
(L)
=⇒ f (L) = 0
That is, the limit L is a root of the function f (x).
2. The tangent line at
x
n
, f (x
n
)
forms a right-triangle with
base x
n
− x
n+1
and height f (x
n
), from which its slope is
f
′
(x
n
) =
f (x
n
)
x
n
− x
n+1
Rearranging this gives the formula x
n+1
= x
n
−
f (x
n
)
f
′
(x
n
)
.
x
y = f (x)
x
n
x
n+1
L
x
n
, f (x
n
)
Newton’s method is particularly nice for polynomials with integer coefficients, since the iterates
form a sequence of rational numbers. This approach was often used obtain rational approximations to
irrational numbers before the advent of calculators.
Examples 4.9. 1. To find a root of f (x) = x
4
+ 4x −6, start with x
0
= 2 and iterate
x
n+1
= x
n
−
x
4
n
+ 4x
n
−6
4x
3
n
+ 4
=
3(x
4
n
+ 2)
4(x
3
n
+ 1)
which yields the sequence (to 3 d.p.)
2,
3
2
,
339
280
, . . .
= (2, 1.5, 1.211, 1.121, 1.114, 1.114, . . .)
You can check with a calculator that 1.114 is approximately
a root.
1 2
x
x
1
x
2
2. The irrational number x =
√
2 +
√
3 is a root of the polynomial
f (x) = x
4
−10x
2
+ 1
By applying Newton’s method with x
0
= 3, we obtain the sequence (to 3 d.p.)
x
n+1
= x
n
−
x
4
n
−10x
2
n
+ 1
4x
3
n
−20x
n
=
3x
4
n
−10x
2
n
−1
4x
n
(x
2
n
−5)
=⇒ (x
n
) =
3,
19
6
= 3.167, 3.147, . . .
51
Newton’s method can be attempted for any differentiable function, though the sequence isn’t guar-
anteed to converge: see for instance Exercise 5. You can find graphical interfaces online for this (for
instance with Geogebra).
Exercises 4.3. 1. Use Newton’s method to find a root of the given function to 4 decimal places.
(Use a calculator, but explain what you are doing!)
(a) f (x) = x
3
−4 (b) f (x) = 2x
3
+ x − 1 (c) f (x) = e
x
−
√
x −2
2. Use Newton’s method to find a rational number approximation to
3
√
2 in lowest terms
p
q
where
10 < q < 100.
3. Suppose you perform Newton’s method for the function f (x) = x
2
− 2 starting with some
positive x
0
> 0.
(a) If x
n
> 0, show that x
n+1
−
√
2 =
1
2x
n
(x
n
−
√
2)
2
=
1
2
−
1
2
√
2x
n
(x
n
−
√
2) .
(b) Explain why
x
n
−
√
2
<
1
2
n
x
0
−
√
2
. Hence conclude that the sequence of iterates (x
n
)
converges to
√
2.
4. We might consider a method of the mean for approximating
3
√
2: given x
0
, define
x
n+1
=
1
2
x
n
+
2
x
2
n
(a) If the sequence (x
n
) converges, show that its limit is
3
√
2.
(b) If x
n
>
3
√
2, show that
2
x
2
n
<
3
√
2.
(c) Let x
0
= 1. Compute x
1
and x
2
. Compare these with the values obtained using Newton’s
method for the function f (x) = x
3
−2 with the same initial condition x
0
= 1.
5. Let f (x) = x
3
−5x.
(a) What happens if you apply Newton’s method to this function with initial condition x
0
=
1? Draw a picture to illustrate.
(b) (Just for fun!) Investigate what happens for other values of x
0
. Can you make any conjec-
tures? Is is possible for x
0
to be positive and yet for x
n
→ −
√
5? Can you make any sense
of what happens if 1 < x
0
<
q
5
3
?
52
