1. No category

Homework 4
Conjecture 1. The nth polygonal number with k sides is given by
the formula
n((k − 2)n + (4 − k))
pn =
.
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We will start by trying to find a recursive formula for the polygonal numbers.
If we find one then we can plug the formula from Conjecture 1 into it and test
whether it holds true.
We can look at a few cases in order to establish a pattern with which to
craft a recursive formula. Looking at the square numbers, we go through the
cases for n = 1 through n = 4. This is shown in Figure 1.
Figure 1: Square Numbers
In the diagram I have set off each additional level of points with dashed
lines and grouped each additional group with rounded rectangles. From these
groupings we can see a pattern emerge in the number of points added as n
increases. For n = 4 we add two groups of three points, plus an additional point
to make seven total. To generalize this, as n increases we add two groups, plus
one point, where each group is composed of n − 1 points.
Our second case will deal with pentagonal numbers for the same set of cases,
using the same procedures for marking the diagram. This case is shown below
in Figure 2.
This example appears to fit the pattern established in the first example. We
again are adding groups of n − 1 points plus one additional point. However, this
time we add three groups instead of two. This is because we have increased the
number of sides in our shape from four to five. If this pattern continues, then
we can say that the recurrence relationship for these polygonal numbers, where
k is the number of sides in the polygon, is
pn = pn−1 + (k − 2)(n − 1) + 1.
Thinking about this geometrically we can confirm that this recurrence relation should work. If we wished to count the number of points making up
the outside edge of each shape, then we would observe that each side is made
of n points. However, merely multiplying this number by the number of sides
would give us a result which was too large. This is because we would be double
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Figure 2: Pentagonal Numbers
counting each of the corners. One way to deal with this issue is to think of
each side as containing n − 1 unique points. Visually we are assigning one of
the endpoints to the first side, and the other endpoint is used in the next side.
Thus each shape is made of k(n − 1) points.
However, we already have a smaller shape providing some of the initial
points, so we will not be adding all k(n − 1) points when creating the nth
polygon. Looking at the diagrams, we can see that the previous shape provides
the unique points of one side, and all but one of the unique points for another
side. Thus we must add one point to complete that second side, plus the n − 1
unique points for the remaining k − 2 sides. This confirms that we must add
(k − 2)(n − 1) + 1 points to get the nth polygon from the (n − 1)th polygon.
Now that we have established the recurrence relationship, we wish to apply the equation from Conjecture 1 to see if it holds within this recurrence
relationship. The algebra for this test proceeds below.
pn = pn−1 + (k − 2)(n − 1) + 1
n((k − 2)n + (4 − k))
= (n − 1)((k − 2)(n − 1) + (4 − k))/2 + (k − 2)(n − 1) + 1
2
(n − 1)((k − 2)(n − 1) + (4 − k)) + (2k − 4)(n − 1) + 2
=
2
=
=
(n2 − 2n + 1)(k − 2) + (4 − k)n − (4 − k) + (2k − 4)n − (2k − 4) + 2
2
(k − 2)n2 + (−2k + 4)n + (k − 2) + (4 − k)n − (4 − k) + (2k − 4)n − (2k − 4) + 2
2
=
(k − 2)n2 + (−2k + 4 + 4 − k + 2k − 4)n + (k − 2 − 4 + k − 2k + 4 + 2)
2
=
(k − 2)n2 + (4 − k)n
2
2
n((k − 2)n + (4 − k))
2
Thus we have manipulated the right side of the equation so that it looks like
the left side of the equation. This confirms that indeed our absolute equation
holds for the recurrence relationship. If we look at the base case of n = 1, then
we can see that for all polygons we will have the 1st polygonal number be 1,
which we know to be true. This works like a proof by induction. We have shown
the 1st case to be true for the equation, we have also shown that it proceeds
properly from n−1 to n, so we can say that the equation works for all polygonal
numbers.
Next we wish to look at the recursive equations for the pyramidal numbers,
which are a three dimensional version of the polygonal numbers consisting of the
different levels of a polygonal number stacked upon itself. Look at the Figure
3 for an example. Creating a recursive equation for this is actually quite easy
considering we have already worked out the polygonal number formulas. For
the triangular pyramidal numbers we start with a base case of the single point
of the triangular number where n = 1. When n = 2 for the triangular pyramidal
numbers, we merely slide the triangular number where n = 2 below the previous
level of triangular pyramidal numbers. For n = 3 we take the current shape and
add the n = 3 case of the triangular numbers.
=
Figure 3: Triangular Pyramidal Numbers, n = 1 through n = 3
If we designate Pn as the nth pyramidal number, then recursively we define
Pn as
Pn = Pn−1 + pn .
However, we know a general formular for pn , which can be substituted into
our recursion equation to get
n((k − 2)n + (4 − k))
,
2
where k is the number of sides in the polygonal number.
If we were to extend this concept to figurate numbers, which are Polygonal
or Pyramidal numbers but with higher dimensions, then we could again find a
Pn = Pn−1 +
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recursive equation. An example to examine would be creating a fourth dimensional figurate number made up of triangular pyramidal numbers. The first level
would only contain the first level of triangular pyramidal numbers. The next
level would extend into the fourth dimension with the second level of triangular
pyramidal numbers. Generalizing this, we get the recursive equation for these
figurate numbers, denoted as Fnd , where d is the dimension we are examining,
d
Fnd = Fn−1
+ Fnd−1 .
This is now our general form for the recursive relationship of all figurate
numbers.
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