Stellar Numbers Project

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COLEGIO ANGLO COLOMBIANO

Stellar Numbers

InvestigationIB Standard Level MathematicsJuan Pablo Gonzlez

13/02/2012


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Introduction

This is a project for the Standard Level Mathematics subject adjourned to the

International Baccalaureate Diploma Programme, in which we will complete several tasks,

all related to certain geometric shapes that can lead and be related to special numbers. For

this purpose, we will use two applications other than Microsoft Word. The first is a webapplication called Wolfram Alpha, which is a computational knowledge engine that helps

us graph the equations that will be explained further on. The second is the native Mac OS X

application Grapher, which will also serve this purpose. Images of the patterns will be used

and will be credited as appropriate

Part ITri angular Numbers

This first part will consider triangular numbers. Triangular numbers can be

represented as triangular patterns of evenly spaced dots, whose width increases as we addrows of dots, as shown on the diagram below:

Our first task is to complete the sequence displayed in the diagram. This is shown in the

diagram below. The diagram contains the triangles corresponding to T6-T8.


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The patterns in the image above provide information with which we can create a table. The

table allows us to determine the nature of the function pertaining to the patterns, so that we

can employ mathematical skills to work out the function itself. Because of this, our table

will include the first and second differences pertaining to the patterns.

n 1 2 3 4 5 6 7 8

y 1 3 6 10 15 21 28 36

table responds to is a quadratic function. Thus, we will have to find an equation with the

formy=ax2+bx+c.

For this purpose we will use a system of simultaneous equations, using the first three n and

y values displayed in the table. Before this, it is important to note that T0is the triangle with

no points, and thus has coordinates (0,0), making c in the equation equal to 0.

Equation 1 (taken from (1,1)):

y = a(1)2+b(1)+c

1 = a+b

Equation 2 (taken from (2,3)):

y = a(2)2+b(2)+c

3 = 4a+2b

Equation 3 (taken from (3,6))

y = a(3)2+b(3)+c6 = 9a+3b

With these three equations, we can work

out both a andb.

E2E1:

4a+2b=3

2(a+b=1)

Thus:

4a+2b=3

-2a+2b=2

2a=1

a=0.5

E3E2:

4(9a+3b=6)

9(4a+2b=3)

Thus:

2 3 54 76 8

1 1 1 11 1

First

diff.

Second

diff.


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36a+12b=24

36a+18b=27

-6b=-3

b=0.5

We thus have found that in the function pertaining to the triangular numbers pattern, both a

and b equal 0.5. This allows us to work out the function itself as follows:

y = n2+n

y=(n2+n)/2

We now have to prove that this function is correct and does indeed work. For this, we will

use T3:

y = (32+3)/2

y = (9+3)/2

y = 12/2

y = 6 Function found is valid.

To further prove the validity of this formula, we employed a CasioFX-9750GIIGD

calculator, and used its graphing function to display the graph for the formula we found.

We then proceeded to use they-calc function, which finds they value for any givenx value

of the function being displayed. We entered all thex values displayed on the table in this

document, and the calculator returned the propery values. The graph for the formula will

be included below:

Part I IStell ar Numbers


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We must now consider stellar numbers (star shapes) withp-vertices. For

mathematical study purposes we will start by looking at 6-stellar shapes, of which the first

four stages are shown in the diagram below:

Having worked out the first and second differences, we have found the number of dots in

each stage up to S6, and we have displayed the data in the table below, which displays those

differences, again serving the purpose of helping us identify, once again, the type of

function related to stellar numbers.

n 0 1 2 3 4 5 6

Sn 0 1 13 37 73 121 181

Once again, we notice that the second difference is always the same, thus indicating that,

akin to the triangular numbers, stellar numbers are defined by a quadratic function. Before

we go into finding the general statement for 6-stellar shapes, we will use the data on the

table above to work out the number of dots in stage S7:

-We know that the first difference between S6 and S7 will be 12 more than the difference

between S5 and S6. This means that the difference will be 60+12=72.

-This means that we will have to add 72 to they value forS6: 181+72=253. The expression

for this will be S7= 12(T6) + 1

12 24 4836 60

12 12 1212

First

diff.

Second

diff.


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Now that we have done this, we will proceed to find the general statement for 6-stellar

shapes, albeit with a different method. We will identify and use the patterns within the

numbers to make this task easier. We have managed to identify the following important

patterns:

The values fory are essentially multiples of 12 to which we add 1 (the centre pointin each shape). This is related to the second difference, which is constantly 12.

12 is the number of vertices (note that a vertex is a corner formed by two edges ofthe shape) in a 6-pointed star, which is in question in this particular task.

The first pattern that was observed and pointed out lets us write the stellar numbersas follows:

- 12(0)+1=1

- 12(1)+1=13- 12(3)+1=37

- 12(6)+1=73

- 12(10)+1=121

- 12(15)+1=181

This leads us to the next pattern: the co-factors of 12 (0, 1, 3, 6, 10 and 15) are infact the triangular numbers we observed in Part I. This leads us to a crucial factor:

stellar numbers are in fact functions of triangular numbers. To elaborate on this, we

will denote the following condition.

-S1 (the first stellar number) is a function of T0 (the zeroth triangular number), just as

S2 is a function of T1.

This means that Sn is a function of Tn-1. Thus, if Tn is equal to (n2+n)/2, then:-Tn = n/2 * (n+1)

-Tn-1 = (n-1)/2 * (n+1-1)

= (n-1)/2 * n

= (n2-n)/2

We then can assume that ifTn-1=(n

2

-n)/2, and Sn = 12(Tn-1) + 1, thenSn = 12((n2-n)/2) + 1

After which it is a matter of mere expansion to work out the general statement:

Sn = 12 * (n2n)/2 +1

= (12n212n)/2 +1

= 6n26n + 1


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We must now test the validity of our statement, using the values in the table we initially

produced. For this we will employ regular mathematical solving:

Sn = 6n26n + 1

= 6(5)26(5) + 1

=6(25)

30 +1=15030 + 1

=121

This was again confirmed by using the graphing function on the GDC model denoted

before, yielding satisfactory results. The graph of the function is again displayed below:

Part II IOther values of P

Part II was completed specifically for 6-stellar shapes (12 vertices, 6 points). Part IIIwill concentrate on 5- and 7-stellar shapes, completing a similar procedure to work out the

general statement for these values ofp.


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5-Stellar Shapes

Below are the first four stages of the 5-stellar shapes:

Again, we have worked out the number of dots in each stage up to S6 by working out the

first and second differences. The data resulting from the procedure is displayed in the table

below:

n 0 1 2 3 4 5 6

Sn 0 1 11 31 61 101 151

Next, we use the data above to work out the number of dots at S7:

-The difference between S6 and S7 will be 50+10=60. Thus, we must add 60 to 151, which

yields the result 211. S7 then has 211 dots in its shape. The expression for this will be S7=12(T6) + 1

We now must find the general statement for 5-stellar numbers. Looking at our work with 6-

stellar shapes, we are able to point out exactly the same set of patterns observed therein.

10 20 4030 50

10 10 1010

First

diff.

Second

diff.


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Taking that into account, we can write the values of Sn as follows:

- 10(0) + 1 = 1

- 10(1) + 1 = 11

- 10(3) + 1 = 31

- 10(6) + 1 = 61

- 10(10) + 1 = 101

- 10(15) + 1 = 151

Having observed that 5-stellar shapes seemingly follow the patterns observed for 6-stellar

shapes, it is safe to conclude that, again, Sn is a function ofTn-1. The only difference would

be that instead of the factor ofTn-1being 12, it is 10, in the case of 5-stellar shapes.

Again, it is then a matter of mere expansion to work out the general formula:

Sn = 10 * (n2-n)/2 + 1

= (10n210n)/2 + 1

= 5n25n + 1

We will confirm the validity of this formula as follows, using the coordinates (5,101):

Sn = 5(5)25(5) + 1

= 5(25)25 + 1

= 12525 + 1

= 101

The graph for the formula is displayed below:


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This equation is essentially a stretched version of the equation found for 6-stellar shapes, as

observed in the graph.

7-Stellar Shapes

We now proceed to repeat the steps already done for 5- and 6-stellar shapes, now

for 7-stellar shapes, of which the first four are displayed below:

The values we were able to work out for stages 5 and 6, along with the first and second

differences for this value ofp are shown on the table below:

n 0 1 2 3 4 5 6

Sn 0 1 15 43 85 141 211

14 28 5642 70

14 14 1414

First

diff.

Second

diff.


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Working out for S7:

- Difference between S6 and S7 will be 70+14=84

- The value of S7 will equal 211 + 84 = 295

- The expression for this will be S7= 12(T6) + 1

Akin to the patterns found for 5- and 6-stellar shapes, we have found that the values for S at

any particular stage are the addition of a multiple of 14, according to the triangular pattern,

to 1. Thus, we can repeat our process of writing the values as such:

- 14(0) + 1 = 1

- 14(1) + 1 = 15

- 14(3) + 1 = 43

- 14(6) + 1 = 85

- 14(10) + 1 = 141

- 14(15) + 1 = 211

Having observed that 7-stellar shapes seemingly follow the patterns observed for 5- and 6-

stellar shapes, it is safe to conclude that, again, Sn is a function ofTn-1. The only difference

would be that instead of the factor ofTn-1being 12 or 10, it is 14.

We thus resort to simple expansion ofSn = 14((n2-n)/2) + 1 to work out the general

statement:

Sn = 14((n2-n)/2) + 1

=(14n214n)/2 + 1

= 7n2

7n + 1

To again test the validity of this statement, we will replace n with 5. By carrying out the

function, we should get 141 as a result:

S5 = 7(5)27(5) + 1

= 7(25)35 + 1

=17535 + 1

=140 + 1

=141

This formula is thus valid. The graph for it is displayed below.


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Part I VGeneral statement f or p-stellar shapes

After completing the process of finding the general statement for three values ofp

inp-stellar shapes (5, 6 and 7), we must now find the general statement. For this we will

identify a possible pattern in the three statements obtained in Parts II and III:

5-stellar shapes: Sn = 5x25x + 1



The pattern jumps to sight:


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