Homework 4 Conjecture 1. The nth polygonal number with k sides is given by the formula n((k − 2)n + (4 − k)) pn = . 2 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 1 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 + 3 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. 4

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