Download THE REGIONS OF A CIRCLE - National Association of Math Circles

THE REGIONS OF A CIRCLE
—–PRELIMINARY DRAFT—–
LEE WINDSPERGER
Overview. The investigation of patterns and sequences is a topic that is pervasive to many
Math Circle projects. Depending on the style in which the Math Circle is conducted, the
treatment of sequences can either be informal, based on elementary hands-on examples
and activities, or in the form of more traditional enrichment classes preparing students for
extracurricular, competitive activities. This Math Circle project is composed of four modules
that aim to provide participants with a solid mathematical introduction to the analysis of
patterns based on the equally fascinating, beautiful, and useful tools consisting of generating
functions, z-transforms, and the field of sequences (i.e., sequences with standard addition
and multiplication defined via the Cauchy product).
The project attempts to provide this introduction in a careful and purposeful manner
so as to foster the students’ individual sense of discovery and creativity and in doing so,
further develop their mathematical confidence and maturity. To motivate the study of the
field of sequences, generating functions, and z-transforms, many interesting modules would
be suitable (for example, difference equations like generalized Fibonacci sequences an+2 =
Aan+1 + Ban , Catalan numbers (Dyck paths, Diagonal Triangulation of Polygons), Motzkin
paths, Hipparchus problem, etc.). We chose to start our first module with the famous
“Regions of a Circle” problem that leads to the analysis of the sequence
1, 2, 4, 8, 16, 31, 57, 99, 163, 256, · · · .
The investigation of patterns in Math Circle type settings allows for creativity and exploration and is a prime example of a problem that can and should be treated with a wide
variety of mathematical tools from all fields of mathematics. Not only do we want students
to experience the awe of finding an answer, we want them to see the beauty and mathematical power of finding answers with seemingly entirely different approaches and methods.
Although some gifted students may be able to have significant success with pattern problems
based solely on their own abilities and ingenuity, the purpose of our second module is to provide a template for a mathematically solid introduction to some of the standard tools that
may be useful when investigating patterns: the field of sequences, generating functions, and
the z-transform. Finally, our third module is dedicated to explorations on how these tools
can be used to give a broader base of students the ability to experience personal success with
the investigations of discrete patterns and how the new approach can be used effectively in
tandem with other more geometrical and graph-theoretic methods that are often used in the
study of patterns.
1
Part 1: Regions of a Circle. This introductory module will be an open investigation of
the following problem:
Given a circle with n points on its boundary, join all pairs of these points with straight
lines. What is the maximum number of regions formed by the lines inside the circle?
We were originally introduced to this problem from Timothy Gowers book [1] and later
discovered that James Tanton had already written an outline for a math circle activity
centered around the same problem which can be found in the lesson plans section of the
mathcircles.org website (see http://www.mathcircles.org/node/670 or [6]). For this module,
we are less interested in leading the students through a presentation similar to Tanton’s
to derive a formula for the number of regions using a strictly combinatorial argument (of
course, there will be no complaints if gifted students are able to produce such an argument
independent of guidance). Rather, we are most concerned with the students developing the
sequence from which we can motivate the introduction of generating functions in the next
module.
A natural starting point for the session is to simply present the regions of the circle problem
and then ask the students to produce the answers for the n = 1, 2, 3, 4, 5. This leads to the
first surprise.
(a) 1 region
(b) 2 regions
(c) 4 regions
(d) 8 regions
(e) 16 regions
Using this data, one can pose the following question: “Can anyone make a conjecture
about the number of regions for n = 6 or for any n?” It is natural to expect that someone
will propose that for any n, the number of regions created by the lines joining n pairs of dots
will be 2n−1 . But could this conjecture be correct?
It will not be very difficult for the students to discover our initial conjecture fails for
n ≥ 6. Besides just simply calculating that for n = 6 the number of regions created is 31
(and not 32 as the initial conjecture states), it can be beneficial to encourage the students
to develop further arguments to critique our initial conjecture. As n gets larger one could
argue that 2n−1 may grow too fast to represent the number of regions. This can simply be
done pictorially. Does it makes sense that in the representations below for n = 9 there are
28 = 256 regions...for n = 13 that there are 212 = 4096 regions? As argued by Gowers in [1]
for n = 30, the 2n−1 hypothesis would estimate over five hundred million different regions.
One could imagine (or actually make) a circle with a diameter of ten meters in a field and
then place thirty pegs around the diameter, connecting all pairs of pegs with strings. As
tedious as it may be to count the regions created by the intersection of these strings, it could
be done. If the hypothesis were true, then there would have to be an average of six hundred
regions per square centimeter, a number which seems to be much too large. Of course, this
argument is by no means mathematically sound and these arguments are only mentioned
with the hope that it encourages the students to reason practically, as well as abstractly,
when they are critiquing and creating mathematical arguments.
2
(f) n = 9
(g) n = 13
This point in the module can be a nice jumping point to tangential explorations of other
surprising sequences. One such example, which Dr. Daniel Ullman introduced to me, is the
sequence associated with Conway’s solitaire army. The n-th term of this sequence is equal to
the minimum number of pegs that can be placed in any number of holes at the integer lattice
points in the lower half-plane in order to advance at least one peg with standard peg solitaire
jumping rules n units into the upper half-plane. Starting with n = 0, the sequence begins
with 1, 2, 4, 8, what is the next term? The answer turns out to be 20, but the bigger surprise
is the next term of the sequence, which is infinity (such a move is impossible). One could
also present historical examples of statements which are true for the first ten, or thousand,
or even million natural numbers but in general prove to be erroneous. One such example
(see [5]), follows from the numbers
0
1
2
3
4
22 + 1 = 3, 22 + 1 = 5, 22 = 17, 22 + 1 = 257, 22 + 1 = 65537
which are all prime numbers. In the 17th century, mathematician Pierre de Fermat claimed
n
that 22 +1 must be a prime number for every positive integer n. A century later, Leonhard
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Euler refuted Fermat’s claim by showing that 22 + 1 = 4294967297 = 641 · 6700417. The
second example (also from [5]) can be seen if one evaluates the expression 991n2 + 1 for
small values of n. For “small” values of n the resulting number is not the square of a whole
number, but for
n = 12055735790331359447442538767
the value is a perfect square. Such examples can lead to a great discussion of mathematical
induction and the care needed to properly construct such a proof.
Refocusing on the original regions of a circle program, the remainder of the exploration
should encourage the students to compute as many terms of the sequence as possible. One
approach is to use technology. Programs such as Mathematica can easily be used to model
the regions of the circle problem. The following is an example of such a code:
Graphics[Table[{Circle[{0, 0}, 1], Line[{{cos[i ∗ 2 ∗ Pi/n], sin[i ∗ 2 ∗ Pi/n]},
{cos[k ∗ 2 ∗ Pi/n], sin[k ∗ 2 ∗ Pi/n]}}]}, {i, 0, n}, {k, 0, n}]].
When students begin to write such a program, they must decide how to arrange the points
on the circle. Should they be placed randomly? Should one allow for curved lines? If one
chooses to place the points evenly spaced on the circle, the representations are symmetric.
One could argue that these representations are perhaps more aesthetically beautiful, and
3
furthermore, attention to symmetry can help students compute higher terms more easily
as well as present further avenues for inquiry and discovery. Below are representations for
n = 6, 7, . . . , 13 created with the above Mathematica code:
(h) n = 6
(i) n = 7
(j) n = 8
(k) n = 9
(l) n = 10
(m) n = 11
(n) n = 12
(o) n = 13
From these representations, many basic questions and problems follow (encourage the
students to ask these questions): Can one prove for which values of n is there an intersection
of lines occurring at the center of the circle? For which values is there no intersection point in
the center of the circle? What happens if the program draws more than two lines intersecting
at the same point...how does this affect the number of regions? If n is odd, do we ever have
more than two lines intersecting at the same point? These types of questions can lead to
good discussions as well as prepare the students for calculating higher order terms in the
“regions of a circle” sequence. Looking at the representations for large n such as n = 22 and
n = 27 could prove to be interesting “math art” for a classroom or bedroom wall.
(p) n = 22
(q) n = 27
4
After producing, digesting and discussing the representations, the students are now ready
to use the symmetries to more easily count the number regions for higher n. For example,
consider n = 9.
(r) n = 9
(s) n = 11
From the Mathematica program’s representation (with regions colored using Microsoft
Paint), one can see that there are 9 yellow pie pieces (each with 11 regions), 9 light blue
quadrilaterals (each with 5 regions), 9 red triangles, 9 dark blue triangles, and 1 white center
region, which gives a total of 163 regions for n = 9. For n = 11, there are 11 yellow pie pieces
(each with 23 regions), 11 light blue quadrilaterals (each with 9 regions), 11 red triangles,
11 dark blue triangles, 11 purple triangles, and 1 white center region, which gives a total
of 386 regions for n = 9. Further questions could be asked about the sequence generated
from counting the number of regions in only the yellow pie piece shapes (or only the blue
quadrilateral shapes). Can we predict how this sequence will grow as we increase n?
For even n, similar arguments can be made, but one must be careful with the intersection
points of more than two lines. For n = 8, observe the following representation.
(t) n = 8
5
As depicted in the figure, there are 8 yellow pie pieces each with 11 regions. There are
also 8 points where three-lines intersect and 1 point where four-lines intersect. The students
should be able to argue that there will be 1 additional region created by each of the three-line
intersection points and 3 additional regions created by the four-line intersection point. Thus
for n = 8, there are 8(11) + 7 + 3 = 99 regions. What about an intersection point of five lines
or six lines or n lines? The students should attempt to argue, in general, that any n-line
additional regions. An understanding
intersection should account for potentially (n−1)(n−2)
2
of how the number of regions is affected by points of intersections is crucial to counting the
number of regions in the following representations:
(u) n = 10
(v) n = 12
Beyond the allure of using technology and creating nice pictures, the hope of presenting
these counting techniques using symmetries is to engage a broader audience than the more
sophisticated, strictly combinatorial approach found in [6]. By the end of the session, the
students should have calculated the number of regions for all n up to 12 and thus generated
the following sequence:
a = (1, 2, 4, 8, 12, 16, 31, 57, 99, 163, 256, 386, 562, · · · ).
In addition, the students will have (hopefully) developed a personal interest in the sequence
and thus be open to learning about different methods to more efficiently calculate higher
order terms of this and other sequences.
6
Part 2: Algebra Approach to Regions of a Circle. The last module led to the study
of a sequence a given by
a = (1, 2, 4, 8, 12, 16, 31, 57, 99, 163, 256, 386, 562, · · · ).
This module assumes that a formula or rule for the nth term of the sequence has not yet
been determined and will work towards such a general formula without assuming any combinatorial intuition. The argument will only require prior knowledge of solving systems of
equations and induction. The methods introduced in this section can be used to solve the
general formula of any sequence whose terms exhibit polynomial growth. The next module
introduces generating functions which can be used to study all sequences.
In this module, the students will turn their attention away from the pictorial representations of the regions of the circle problem and focus the investigation on what is known
of the sequence a. Instead of solely looking for hidden clues and patterns within a, which
the students have probably already attempted in the first module, one should encourage the
students to look at the behavior of the sequence d1 of the changes (differences) of the terms
of a. This can serve as a nice philosophical lesson for the students: in working toward an
understanding of the true nature of something, it is often helpful to understand how that
something changes. Thus consider the sequence
d1 = 2 − 1, 4 − 2, 8 − 4, 16 − 8, 31 − 16, 57 − 31, 99 − 57, 163 − 99, 256 − 163,
386 − 256, 562 − 386, · · ·
1
d = 1, 2, 4, 8, 15, 26, 42, 64, 93, 130, 176, · · · .
Nothing should be too striking after investigating d1 , but encourage the further investigation of d2 (the differences between consecutive terms of d1 ), d3 (the differences in d2 ), and
dn for n > 3:
d2 = 1, 2, 4, 7, 11, 16, 22, 29, 37, 46, · · ·
d3 = 1, 2, 3, 4, 5, 6, 7, 8, 9, · · ·
d4 = 1, 1, 1, 1, 1, 1, 1, 1, 1, · · · .
Now the students should witness the first revelation of this module, the fourth differences
of the regions of a circle sequence a are constant. Of course, at this stage one can not be sure
that this is true for all terms of a, but it certainly holds for the first few terms. Encourage the
students to move forward with the assumption that the fourth differences of a are constant
for all terms, and in doing so, always emphasize that this assumption has to be eventually
verified in order to fully understand the problem. The students should now be able to give
insight to the type of equation a general formula should be.
In most high school geometry classes, it is taught that if the terms of a sequence can be
represented by a linear function, then the first differences of the terms are constant. Also,
most students have seen that, if the general formula for a given sequence is quadratic, then
the second difference of the terms must be constant. Ask the students to make a conjecture
about what type of equation the general formula of a sequence should be if the pth difference
of that sequence is constant. A reasonable conjecture follows:
Given a sequence f whose general formula f (n) for the nth term f is a polynomial of
degree p, then the pth difference of the terms of f must be constant.
7
The students should be able to produce or follow an inductive argument for this claim.
If f is a polynomial of degree 0, then f must be a constant sequence given by f (n) = c.
Clearly, the difference between the terms of f is 0. If f is assumed to be a linear function
(polynomial of degree 1) then f (n) = an + b for some constant a and b. Observe that
f (n + 1) − f (n) = a(n + 1) + b − (an + b) = a,
which proves that the first difference of the terms of f is constant. For the inductive hypothesis, assume that the claim holds for polynomials of degree p − 1. Let f be a sequence
whose general formula is a polynomial of degree p. Then
f (n) = anp + bnp−1 + cnp−2 + . . . + d
f (n + 1) − f (n) = a(n + 1)p + b(n + 1)p−1 + c(n + 1)p−2 + . . . + d
−(anp + bnp−1 + cnp−2 + . . . + d)
= polynomial of degree p − 1.
By the inductive hypothesis, the p − 1 difference of the sequence determined by taking the
first difference of f is a constant. Thus the pth difference of f is a constant, and the claim is
verified. The circle could further investigate if an argument can be produced for the converse
of the above claim.
With the above knowledge, the students should now have a candidate for the type of
equation that the general formula for a could be. This equation could be a fourth degree
polynomial, thus one can assume
f (n) = an4 + bn3 + cn2 + dn + e.
Using the first four terms of a, the following system of equations follows:
a(1)
a(2)
a(3)
a(4)
a(5)
=
=
=
=
=
a+b+c+d+e
16a + 8b + 4c + 2d + e
81a + 27b + 9c + 3d + e
256a + 64b + 16c + 4d + e
625a + 125b + 25c + 5d + e
=1
=2
=4
=8
= 16.
This system can be solved by hand or, perhaps more efficiently, by using technology (a
TI-83 graphing calculator works nicely). This means that
a(n) =
1 4
6
23
18
n − n3 + n2 − n + 1
24
24
24
24
is a candidate for a general formula for a. The students should explore whether this formula
holds true for the known values of a. Remarkably, the formula does hold true for all known
values of a, and thus the value of a(n) appears to be an integer for all integer n. What else
can the students learn about the original regions of the circle problem by looking at this
formula? There is little insight that one can draw from this representation of a potential
solution other than the fact that the final solution needs to be an integer polynomial, meaning
that the formula produces integer value outputs for all integer value inputs. Perhaps there
is a more insightful representation for the general formula yet to be discovered?
In searching for this representation, the students should be encouraged to research integer
polynomials. The first result from a google search [7] informs that every integer polynomial
8
f (x) of degree p in the variable x can be written in the form
x
x
x
(0.1)
f (x) = A0 + A1
+ A2
+ ... + Ap
1
2
p
where A0 , A1 , ..., An are integers. The website directs readers to Nagell [4] for the proof of
the claim. The proof in [4] can clearly be presented to or formulated by high school students.
The first step in showing that every integer representing polynomial can be written in the
form (0.1) is to show that every polynomial f (x) of degree p can be written as
x
x
x
(0.2)
f (x) = c0 + c1
+ c2
+ . . . + cp
1
2
p
where ci are uniquely determined constants. Let f (x) be a polynomial of degree zero. Then
f (x) = c0 and the claim holds. One could either finish the proof using the inductive method
or if the students are not familiar with induction, let the students try to prove the theorem
for polynomials of degree one and two before leading them through the inductive step. Let
f (x) be a polynomial of degree one, then
f (x) = c0 + c1 x = c0 + c1 x1 .
Now let f (x) be a polynomial of degree two, then
f (x) = c0 + c1 x + c2 x2
x2
1
1
= c0 + (c1 + )x − 2c2 ( ) − x
2
2
2
1
x(x − 1)
= c0 + (c1 + )x − 2c2 (
)
2 2
1 x
x
= c0 + (c1 + )
− 2c2
.
2 1
2
The students could then explore, further attempting to prove the claim for higher degree
polynomials. For students who have little experience writing proofs, these exercises can be
good practice and can introduce the value in working backwards from the result they want to
prove. If one wanted to use induction, then assume the hypothesis holds for any polynomial
of degree p − 1. Let f (x) be of degree p, then
f (x) = c0 + c1 x + . . . + cp xp
and define g(x) = f (x) − cp p! xp . Then
x
g(x) = f (x) − cp p!
p
x(x − 1) · · · (x − (p + 1))
= c0 + c1 x + . . . + cp xp − cp p!
p!
p
p
= c0 + c1 x + . . . + cp x − cp x − (polynomial of degree p-1).
9
Hence g(x) is a polynomial of degree less than p. By the inductive hypothesis,
x
x
g(x) = b0 + b1
+ . . . + bp−1
.
1
p−1
It follows that f (x) is of form (0.2) since
x
x
x
f (x) = b0 + b1
+ . . . + bp−1
+ cp p!
.
1
p−1
p
After proving that every polynomial can be written in the form (0.2), to show that every
integer representing polynomial can be expressed as (0.1), one can assume that f (x) is an
integer representing polynomial of degree p. Then express f (x) as
x
x
x
f (x) = c0 + c1
+ . . . + cp−1
+ cp
.
1
p−1
p
Notice that f (0) must be an integer. This implies that c0 is an integer. Now suppose that
c0 , c1 ,...,cp−1 are all integers. Then with x = p, the following holds:
p
f (p) = c0 + c1 p1 + . . . + cp−1 p−1
+ cp pp
p
f (p) = c0 + c1 p1 + . . . + cp−1 p−1
+ cp .
By definition, f (p) must also be an integer. Hence cp must be an integer, and by induction,
every integer representing polynomial can be written as (0.1).
It is clear that the desired formula a(n) must be an integer representing polynomial. Hence,
from the brief diversion into the study of integer representing polynomials, one must be able
to write a(n) as follows:
6 3 23 2 18
n
n
n
1 4
n
+ A2
+ A3
+ A4
a(n) = n − n + n − n + 1 = A0 + A1
.
24
24
24
24
1
2
3
4
The students can now use the first four terms of a to easily solve for the above coefficients.
a(0)
a(1)
a(2)
a(3)
a(4)
With this new representation
(0.3)
=
=
=
=
=
1
1
2
4
8
=
=
=
=
=
A0
1 + A1
1 + A2
1 + 3 + A3
1 + 6 + A4
⇒
⇒
⇒
⇒
⇒
A0
A1
A2
A3
A4
=1
=0
=1
=0
= 1.
n
n
a(n) = 1 +
+
2
4
the students are ready to return to the regions of the circle problem’s pictorial representation.
Does the new representation (0.3) still accurately represent the terms of a? How does (0.3)
differ from the previous representation
6
23
18
1
a(n) = n4 − n3 + n2 − n + 1?
24
24
24
24
Most importantly, are the students able to draw insight to the nature of the problem from
the new representation (0.3)? Looking at a circle with n points at its boundary, every two
points determine a line and every four points determine an intersection point of two lines.
10
Further consideration shows that there are n2 lines and n4 points of intersection. Hence,
(0.3) should illustrate to the students that the maximal number a(n) of regions in the circle
produced by connecting n points on the boundary with lines should be identical to one plus
the number of lines plus the number of points of intersection of those lines; hence
(0.4)
a(n) = 1 + Ln + In
where Ln = n2 (the number of lines) and In = n4 (the number of intersections) Finally,
the students have a conjecture that they should be ready to prove.
Theorem. In any circle with n points on the boundary, the maximal number a(n) of different
regions that can be generated by connecting the n points with lines is given by a(n) = 1+Ln +
In , where Ln is the number of lines that can be drawn and In is the number of intersections
that can be obtained
Proof. The proof will follow as written in [6]. The equation a(n) = 1 + Ln + In holds
for n = 1. In this case, there is one point, one region, and L1 = I1 = 0. To finish the
proof, it will be shown that the equation remains valid as one adds new points and the lines
connecting the new points to the existing points. Given a new point, as one draws the new
line toward an existing point, each time the new line intersects a preexisting line the number
of intersections increases by one and number of regions increases by one. Hence the equation
remains unchanged. Once the new line reaches a preexisting point then the number of lines
Ln has increased by one. But, as that new line reaches a preexisting point, that line splits
the final region in two. This increases the number of regions by one and thus the equation
remains balanced. This means that the process of increasing the number of points on the
circle’s boundary and adding the lines to connect that point to all the preexisting points
does not change the formula a(n) = 1 + Ln + In .
Now the students have proved that the general formula for the sequence a is given by
n
n
a(n) = 1 +
+
.
2
4
At this point the session can be extended using [6] to explore other combinatorial and graph
theoretic generalizations of the result. Alternatively, the session could also be extended by
having the students explore integer representing polynomials. What do the graphs of these
integer polynomials look like? How does one create a sequence that coincides with the powers
of 2 up to 2n for any n? If the students enjoy working with sequences, then one also has the
option to move to the module in the next section. This module introduces the z-transform
which serves as a useful tool for the study of all sequences. The last module (Part 4) returns
to the regions of the circle sequence an and uses the z-transform to find a general formula
for an .
11
Part 3: Algebra with Sequences. This module aims to emulate but the approach taken
in the previous module, but in a more general setting. This module will be more instructive
in nature, but the material should not be surprising to students with a basic knowledge of
algebra. The goal of the module is to introduce the algebra of sequences. In doing so, the
basic concepts of algebra should be reinforced while, for perhaps the first time, the students
will work with a field type structure other than the rational numbers. For the sake of brevity,
the following material is presented in an instructive manner. Many of the concepts can and
should be presented in an open fashion, allowing the students to discover the results with
their own arguments. For instance, addition is defined below in such a way that the standard
rules of algebra (associativity, commutativity, ... etc) hold. One could ask the students to
discover this definition and offer arguments for why the standard rules hold. Likewise, once
the definition of addition is established, the students should be left with the task of finding
the zero and additive inverse for the operation.
In order to establish the algebra of sequences, the sequence a, and all other sequences to
be considered, must be defined in “both directions” as follows:
a = (· · · , 0, 0, ak , ak+1 , ak+2 , · · · ),
where k ∈ Z and aj = 0 for all j < k. For example, the sequence a considered in the previous
section is written as
a := (· · · , 0, 0, 0, 1, 2, 4, 8, 16, 31, 57, 99, 163, 256, 386, 562, · · · ),
where ak = 0 for integers k ≤ 0, a1 = 1, a2 = 2, a3 = 4, etc. In order to produce a “formula”
for the members ak of the sequence a it comes in handy if one knows how to add and multiply
sequences so that the standard rules of algebra remain valid; that is, one has to know how
to add and multiply sequences a, b, c such that the following properties hold:
name
addition
multiplication
associativity
(a + b) + c = a + (b + c)
a ∗ (b ∗ c) = (a ∗ b) ∗ c
commutativity
a+b=b+a
a∗b=b∗a
distributivity a ∗ (b + c) = a ∗ b + a ∗ c (a + b) ∗ c = a ∗ c + b ∗ c
identity
a+0=a
a∗1=a
−1
inverses
a + (−a) = 0
a ∗ a = 1 if a 6= 0
Fortunately, the addition and multiplication of sequences is not difficult to learn. The
addition of sequences is particularly easy since this is done coordinatewise; i.e.,
k
ak
bk
(a + b)k
···
···
···
···
−5
−4
−3
−2
0
0
0
0
0
0
0
0
0
0
0
0
···
2 1 -1 1 -2 1 0 0 0 0 · · ·
-2 0 -1 0 2 3 0 0 0 0 · · ·
0 1 -2 1 0 4 0 0 0 0 · · ·
−1
0
1
2
3
4
5
6
7
8
The “zero” in the field of sequences is the sequence for which ak = 0 for all integers k.
Moreover, for each sequence a = (· · · , 0, 0, ak , ak+1 , ak+2 , · · · ), the additive “inverse” sequence
is given by −a = (· · · , 0, 0, −ak , −ak+1 , −ak+2 , · · · ).
12
Perhaps the most difficult aspect about working with sequences is keeping track of the
position at which each member of the sequence is located. For this reason, the following
bookkeeping device is extremely useful. When looking at the sequence
k
ak
···
···
−5
−4
−3
−2
−1
0
0
0
0
2
···
1 -1 1 -2 1 0 0 0 0 · · ·
0
1
2
3
4
5
6
7
8
a great book-keeping device is to think of the sequence a as the function
a(z) = 2z −1 + 1 − z + z 2 − 2z 3 + z 4 .
Working in the opposite direction, the sequence
k
bk
···
···
−5
−4
−3
−2
0
0
0
0
···
-2 0 -1 0 2 3 0 0 0 0 · · ·
−1
0
1
2
3
4
5
6
7
8
can be thought of as a bookkeeping device to capture the coefficients of the function
b(z) = −2z −1 − z + 2z 3 + 3z 4 .
The standard method of adding functions yields that
a(z) + b(z) = (2z −1 + 1 − z + z 2 − 2z 3 + z 4 ) + (−2z −1 − z + 2z 3 + 3z 4 )
= 1 − 2z + z 2 + 4z 4 .
The sequence notation shows that that the a + b can be written in the following sequence
form:
k
(a + b)k
···
···
−5
−4
−3
−2
−1
0
0
0
0
0
···
1 -2 1 0 4 0 0 0 0 · · ·
0
1
2
3
4
5
6
7
8
This clever way of thinking of sequences is called the z-transform or the method of generating
functions [8]. The z-transform will prove itself to be an extremely powerful tool when working
with sequences. According to Kirk and Strum [2], the z-transform method was used first
by Gardner and Barnes in the early 1940’s to solve linear, constant-coefficient difference
equations and by W. Hurewicz in 1947 to transform a sample signal or sequence. According
to a website maintained by Professor Mark Liberman of the University of Pennsylvania,
the term z-transform originated in 1952 from a sampled-data control group at Columbia
University led by Professor John R. Raggazini [3]. Offering some historical perspective on
the material (especially considering the concept of the z-transform is relatively new) can be
a nice way of showing students that math is an ever changing and growing field of research.
As a first application, the z-transform will serve as a guide in working a good definition for
the multiplication of two sequences. To get started, consider the two sequences
k
ak
bk
···
···
···
−5
−4
−3
−2
−1
0
0
0
0
0
0
0
0
1
0
···
0 0 0 0 0 0 0 0 0 ···
0 1 0 0 0 0 0 0 0 ···
0
13
1
2
3
4
5
6
7
8
If the multiplication of sequences were to be again defined coordinatewise, then a ∗ b = 0,
although a 6= 0 and b 6= 0, which contradicts the rules of algebra. Thus, a more thoughtful
definition of the multiplication of sequences is needed that uses the fact that (using the
z-transform) one can think of the sequence a as the function a(z) = z −1 and of the sequence
b as the function b(z) = z. Since a(z)b(z) = z −1 z = 1, define (a ∗ b)k as follows:
k
ak
bk
(a ∗ b)k
···
···
···
···
−5
−4
−3
−2
−1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
···
0 0 0 0 0 0 0 0 0 ···
0 1 0 0 0 0 0 0 0 ···
1 0 0 0 0 0 0 0 0 ···
0
1
2
3
4
5
6
7
8
Multiplying sequences in this way yields a perfect way to do algebra with sequences. First
of all, the sequence 1 in the field of sequences is the sequence for which ak = 0 for all k 6= 0
and a0 = 1; i.e.,
···
1 ···
k
−5
−4
−3
−2
−1
0
0
0
0
0
···
1 0 0 0 0 0 0 0 0 ···
0
1
2
3
4
5
6
7
8
or 1(z) = 1. In the above example, the sequence a is equal to b−1 since a ∗ b =1. Before
stating the general definition of multiplication of sequences, consider an additional example.
Multiply the following two sequences:
k
ak
bk
···
···
···
−5
−4
−3
−2
−1
0
0
0
0
0
0
0
0
0
0
0
1
2
3
4
5
6
7
a0 a1 a2 a3 a4 a5 a6 a7
b0 b1 b 2 b3 b4 b5 b6 b7
···
···
···
Again, to figure out what a ∗ b might be, consider a, b as
a(z) = a0 + a1 z + a2 z 2 + a3 z 3 + · · · and b(z) = b0 + b1 z + b2 z 2 + b3 z 3 + · · · .
Since
a(z)b(z) = (a0 + a1 z + a2 z 2 + a3 z 3 + · · · )(b0 + b1 z + b2 z 2 + b3 z 3 + · · · )
= a0 b0 + (a0 b1 + a1 b0 )z + (a2 b0 + a1 b1 + a0 b2 )z 2
+(a3 b0 + a2 b1 + a1 b2 + a0 b3 )z 3 + · · · ,
one is led to the definition of how to multiply sequences. Let
a = (· · · , 0, 0, ai , ai+1 , ai+2 , · · · ) and b = (· · · , 0, 0, bj , bj+1 , bj+2 , · · · )
be two sequences. Then the convolution (or Cauchy) product a ∗ b := c, where the k-th
coordinate of c, is defined by
X
ck =
ai b j
i+j=k
or, equivalently, where c(z) = a(z)b(z) = k ck z k . The set of sequences with coordinatewise
addition (+) and multiplication (∗) fulfills all the rules of algebra listed above. Thus, one
can do algebra with sequences in the same manner you do algebra with numbers. Just do it!
P
14
Example. Consider the sequence
i
ai
···
···
−5
−4
−3
−2
−1
0
0
0
0
0
···
1 1 1 1 1 1 1 1 1 ···
0
1
2
3
4
Then
a(z) = 1 + z + z 2 + z 3 + · · · =
5
6
7
8
1
1−z
for |z| < 1 and therefore
1
= 1 − z.
a(z)
is given by
a−1 (z) =
This indicates that the sequence b = a−1
j
bj
···
···
−5
−4
−3
−2
−1
0
0
0
0
0
and it is indeed true (check it!!) that ck =
if k 6= 0.
···
1 -1 0 0 0 0 0 0 0 · · ·
0
1
P
i+j=k
15
2
3
4
5
6
7
8
ai bj is equal to 1 if k = 0 and equal to 0
Part 4: Regions of a Circle Revisited. This module begins similarly to the second module. The investigation returns to the regions of a circle sequence a given by
i
ai
···
···
−5
−4
−3
−2
−1
0
0
0
0
0
···
0 1 2 4 8 16 31 57 99 163 256 386 562 · · ·
0
1
2
3
4
5
6
7
8
9
10
11
12
Again, start with a study of the behavior of the sequence b of the differences (· · · , 0, a1 , a2 −
a1 , a3 − a2 , a4 − a3 , · · · ) or its generating function
b(z) = a1 + (a2 − a1 )z + (a3 − a2 )z 2 + (a4 − a3 )z 3 + · · · = a(z)(z −1 − 1) = (a ∗ d)(z),
where d(z) = z −1 − 1 is the generating function of the sequence
i
di
···
···
−5
−4
−3
−2
−1
0
0
0
0
1
0
1
2
3
4
5
6
7
8
9
10
11
12
-1 0 0 0 0 0 0 0 0 0 0
0
·
···
···
First, what does the sequence b = a ∗ d look like?
i
ai
(a ∗ d)i
−5
−4
−3
−2
−1
0
0
0
0
0
0
0
0
0
0
···
0 1 2 4 8 16 31 57 99 163 256 386 562 · · ·
1 1 2 4 8 15 26 42 64 93 130 176 · · · ·
0
1
2
3
4
5
6
7
8
9
10
11
12
Searching for more insight, the students should investigate the second differences, third
differences, and so forth. Keep in mind that looking at differences of the terms of the
sequence is the same as multipying the generating function of that sequence with the function
d(z) = z −1 − 1 = 1−z
.
z
i
ai
(a ∗ d)i
(a ∗ d2 )i
(a ∗ d3 )i
(a ∗ d4 )i
(a ∗ d5 )i
−5
−4
0
0
0
0
0
0
0
0
0
0
0
1
−3
−2
−1
0
0 0 0
0 0 0
0 0 1
0 1 -1
1 -2 2
-3 4 -2
0
1
0
1
0
1
···
1 2 4 8 16 31 57 99 163 256 386 562 · · ·
1 2 4 8 15 26 42 64 93 130 176 · · · ·
1 2 4 7 11 16 22 29 37 46
·
· ···
1 2 3 4 5 6 7 8
9
·
·
· ···
1 1 1 1 1 1 1 1
·
·
·
· ···
0 0 0 0 0 0 0 ·
·
·
·
· ···
1
2
3
4
5
6
7
8
9
10
11
12
This indicates that the (still largely unknown) generating function a(z) might have the
property that
5
1
1−z
3
4
2
a(z)
= 4 − 3 + 2 − +1
z
z
z
z
z
or
1 − 3z + 4z 2 − 2z 3 + z 4
1
a(z) =
.
z
(1 − z)5
Using the method of partial fractions, there are real numbers A, B, C, D, E such that
1
A
B
C
D
E
a(z) =
+
+
+
+
.
2
3
4
z
1 − z (1 − z)
(1 − z)
(1 − z)
(1 − z)5
16
Comparing the coefficients in the equation z 4 − 2z 3 + 4z 2 − 3z + 1 = A(1 − z)4 + B(1 − z)3 +
C(1 − z)2 + D(1 − z) + E = Az 4 − (4A + B)z 3 + (6A + 3B + C)z 2 − (4A + 3B + 2C + D)z +
(A + B + C + D + E) yields that A = 1, 4A + B = 2, 6A + 3B + C = 4, 4A + 3B + 2C + D = 3,
and A + B + C + D + E = 1 or A = 1, B = −2, C = 4, D = −3, E = 1. This shows that
1
1
2
4
3
1
a(z) =
−
+
−
+
.
z
1 − z (1 − z)2 (1 − z)3 (1 − z)4 (1 − z)5
For |z| < 1 (an opportunity for a review and discussion of the geometric series)
1
= 1 + z + z2 + z3 + z4 + · · · .
1−z
Multiplying a function a(z) = a0 + a1 z + a2 z 2 + a3 z 3 + · · · with
yields the function
a(z)
1
1−z
= 1 + z + z2 + z3 + z4 + · · ·
1
= a0 + (a0 + a1 )z + (a0 + a1 + a2 )z 2 + (a0 + a1 + a2 + a3 )z 3 + · · · .
1−z
Thus,
1
(1 − z)2
1
(1 − z)3
1
(1 − z)4
1
(1 − z)5
= 1 + 2z + 3z 2 + 4z 3 + · · · + (n + 1)z n + · · ·
(n + 1)(n + 2) n
z + ···
2
(n + 1)(n + 2)(n + 3) n
= 1 + 4z + 10z 2 + 20z 3 + · · · +
z + ···
3!
(n + 1)(n + 2)(n + 3)(n + 4) n
= 1 + 5z + 15z 2 + 35z 3 + · · · +
z + ··· ,
4!
= 1 + 3z + 6z 2 + 10z 3 + · · · +
and therefore z1 a(z) =
∞
X
n=0
(1−2(n+1)+4
1
1−z
−
2
(1−z)2
+
4
(1−z)3
−
3
(1−z)4
+
1
(1−z)5
=
(n + 1)(n + 2)
(n + 1)(n + 2)(n + 3) (n + 1)(n + 2)(n + 3)(n + 4) n
−3
+
)z
2
3!
4!
or
∞ X
n
n+1
n+2
n+3
n+4
a(z) =
(
−2
+4
−3
+
)z n+1 .
0
1
2
3
4
n=0
17
Thus an+1 =
n
0
−2
n+1
1
an =
=
=
=
=
=
=
− 3 n+3
+ n+4
for n ≥ 0 or, for n ≥ 1,
3
4
n
n+1
n+2
n+3
1−2
+4
−3
+
1
2
3
4
n+2
n
n+1
n+2
1−2
+4
−2
+
1
2
3
4
n
n+1
n+1
n+2
1−2
+2
−2
+
1
2
3
4
n
n+1
n+1
n+1
1−2
+2
−
+
1
2
3
4
n
n+1
n+1
1+2
−
+
2
3
4
n
n
n+1
1+
−
+
2
3
4
n
n
1+
+
.
2
4
+4
n+2
2
The students have now derived the same candidate for a general formula for the nth term of
the sequence a. At this point one should return to the second module (Part 2) to finish the
investigation and proof of the general formula. After doing so, the students should not only
have a very good understanding of the regions of a circle problem, but also new techniques
for studying and gaining insight to sequences. With these new insights and techniques
the students are now ready to investigate other sequences such as sequences derived from
difference equations like generalized Fibonacci sequences an+2 = Aan+1 + Ban , the Catalan
numbers, Dyck or Motzkin paths, Hipparchus problem, etc.
References
[1]
[2]
[3]
[4]
[5]
[6]
Gowers, Timothy. Mathematics: A Very Short Introduction, Oxford University Press, 2002.
Kirk, Strum Contemporary Linear Systems, BrooksCole, 2000, p 420.
M. Liberman, University of Pennsylvania http://www.ling.upenn.edu/courses/ling525/z.html
Nagell, T Introduction to Number Theory, Wiley, 1951, p 121.
Sury, B. Mathematical Induction - An Impresario of the Infinite, Resonance, February 1998.
Tanton, James. Clip Theory, Mathematical Outpourings: Newsletters and Musing from the St. Mark’s
Institute of Mathematics, 2010.
[7] Weisstein, Eric W. Integer-Representing Polynomial. MathWorld–A Wolfram Web Resource.
http://mathworld.wolfram.com/Integer-RepresentingPolynomial.html
[8] Wilf, Herbert S., Generating Functionology, Academic Press, 1994, pp 1-65.
Department of Mathematics, Louisiana State University, [email protected]
18