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Stellar Numbers. After establishing the general formula for the triangular numbers, stellar (star) shapes with p vertices leading to p-stellar numbers were to be considered.

Extracts from this document...

Introduction

Olivia Bloch        Copenhagen International School        30.05.2011

The aim of this task is to investigate geometric shapes, which lead to special numbers. The simplest example of these are square numbers, such as 1, 4, 9, 16, which can be represented by squares of side 1, 2, 3, and 4.

Triangular numbers are defined as “the number of dots in an equilateral triangle uniformly filled with dots”. The sequence of triangular numbers are derived from all natural numbers and zero, if the following number is always added to the previous as shown below, a triangular number will always be the outcome:

1 = 1

2 + 1 = 3

3 + (2 + 1) = 6

4 + (1 + 2 + 3) = 10

5 + (1 + 2 + 3 + 4) = 15

Moreover, triangular numbers can be seen in other mathematical theories, such as Pascal’s triangle, as shown in the diagram below. The triangular numbers are found in the third diagonal, as highlighted in red.

image00.png

The first diagrams to be considered show a triangular pattern of evenly spaced dots, and the number of dots within each diagram represents a triangular number.

image01.png

Thereafter, the sequence was to be developed into the next three terms as shown below.

image09.pngimage20.pngimage31.png

The information from the diagrams above is represented in the table below.

Term Number (n)

1

2

3

4

5

6

7

8

Triangular Number (Tn)

1

3

6

10

15

21

28

36

Establishing the following three terms in the sequence was done by simply drawing another horizontal row of dots to the previous equilateral and adding those dots to the previous count. However, following the method described earlier can also do this calculation, as shown in the illustration below.

T1        1 = 1

T2        2 + 1 = 3

T3        3 + (2 + 1) = 6

T4        4 + (1 + 2 + 3) = 10

T5        5 + (1 + 2 + 3 + 4) = 15

T6        6 + (1 + 2 + 3 + 4 + 5) = 21

T7        7 + (1 + 2 + 3 + 4 + 5 + 6) = 28

...read more.

Middle

121

181

image06.png

As seen in the diagram above, the second difference is the same between the terms, and the sequence is therefore quadratic. This means that the equation Sn = an2 + bn + c will be used when representing the data in a general formula. Since some of the values of Sn have already been established this makes it possible to work out the general formula. The first step is to substitute the established values into the three quadratic equations, as shown below:

Sn = an2 + bn + c

Therefore: quadratic

When n = 1, Sn = 1

1 = a(1)2 + b(1) + c

1 = 1a + 1b + c

1 = a + b + c

When n = 2, Sn = 13

13 = a(2)2 + b(2) + c

13 = 4a + 2b + c

When n = 3, Sn = 6

37 = a(3)2 + b(3) + c

37 = 9a + 3b + c

Thereafter polysmlt on the graphic display calculator is used in order to retrieve the values of a, b, and c. The coefficients above are simply plugged into the calculator; this can be seen in the screenshots below

image08.pngimage07.png

Hence the general formula for the sequence is Sn = 6n2 - 6n + 1, to be certain that this is in fact correct, it can also be worked out as shown below.

image10.png

The second difference of the sequence is divided by 2, since this is the value of a, thereafter the sequence is extended to T0, as the number highlighted in red is c. To figure out the value of b, the values have been substituted into the quadratic formula as shown below, where after the equation has been solved using simple algebra.

Sn = an2 + bn + c

Sn = 6n2 + bn + 1

S1 = (6)(1)2 + b(1) + 1

1= 6 + b + 1

b = - 6

∴ Sn = 6n2 - 6n + 1

As seen above, this method works out the same general formula, and this is tested below to assess the validity.

Sn = 6n2 - 6n + 1

S3 = 6(3)2 – 6(3) + 1

S3 = 6(9) – 18 + 1

S3 = 54 – 17

S3 = 37

Sn = 6n2 - 6n + 1

S5 = 6(5)2 – 6(5) + 1

S5 = 6(25) – 30 + 1

S5 = 150 – 29

S5 = 121

The results derived from the general formula are the same as worked out earlier when not applying any formula at all; hence, it is correct and can be applied to terms 3 and 5 and also other values of n.

The previous example was then repeated using other values of p; hence, the number of vertices is being changed from 6 to 5. The 5-stellar number at each stage represents the number of dots in each of the diagrams below.

image11.png

image12.png

The information in the diagrams above was collected and is represented in the table below.

Term Number (n)

1

2

3

4

5

6

Stellar Number (Vn)

1

11

31

61

101

151

This information is considered and the pattern in the sequence is represented below, and a quadratic pattern throughout the investigation has become very obvious.

image13.png

As seen in the diagram above, the second difference is the same between the terms, and the sequence is therefore quadratic. This means that the equation Vn = an2 + bn + c will be used when representing the data in a general formula. Since some of the values of Vn have already been established this makes it possible to work out the general formula. The first step is to substitute the established values into the three quadratic equations, as shown below:

Rn = an2 + bn + c

Therefore: quadratic

When n = 1, Vn = 1

1 = a(1)2 + b(1) + c

1 = 1a + 1b + c

1 = a + b + c

When n = 2, Vn = 13

11 = a(2)2 + b(2) + c

11 = 4a + 2b + c

When n = 3, Vn = 6

31 = a(3)2 + b(3) + c

31 = 9a + 3b + c

As seen above, this method works out the same general formula, and this is tested below to assess the validity.

Thereafter polysmlt on the graphic display calculator is used in order to retrieve the values of a, b, and c. The coefficients above are simply plugged into the calculator; this can be seen in the screenshots below

image14.pngimage15.png

Hence, the general formula for this sequence is Rn = 5n2 - 5n + 11

image16.png

Vn = an2 + bn + c

Vn = 6n2 + bn + 1

V1 = (5)(1)2 + b(1) + 1

1= 5 + b + 1

b = - 5

∴ Sn = 5n2 - 5n + 1

Vn = 5n2 - 5n + 1

V3 = 5(3)2 – 5(3) + 1

V3 = 5(9) – 15 + 1

V3 = 45 – 14

V3 = 31

Vn = 5n2 - 5n + 1

V5 = 5(5)2 – 5(5) + 1

V5 = 5(25) – 25 + 1

V5 = 125 – 24

V5 = 101

The results derived from the general formula are the same as worked out earlier when not applying any formula at all; hence, it is correct and can be applied to terms 3 and 5.

This was repeated using yet again another value for p, this time changing the p value from 5 to 4, resulting in the diagrams below.

image17.pngimage18.png

The information from the diagrams above was collected and is represented in the table below.

Term Number (n)

1

2

3

4

5

6

Stellar Number (Rn)

1

9

25

49

81

121

...read more.

Conclusion

image32.png

As seen on the left, the values in the table on the top correspond to the excel document on the bottom. I plugged in the general formula discussed above on the far right hand side, and the outcomes were as displayed on the left. Using other values of p and n, which would allow for a more detailed observation, extended this study, which can be seen below. To ensure that the formula in excel did indeed calculate the correct values, I set up the previous results up to compare, and they did correspond. image35.pngimage34.pngimage33.png

This investigation has considered geometric shapes in order to determine how many dots are in the different types of shapes, in this case from triangular to stellar, furthermore extending the investigatory work by changing the stellar number (the value of p). The limitations of this investigation is that the value of n must always be equal to or greater than 1, as it is impossible to have a negative triangle or stellar shape, as it is a physical entity, and it is therefore also impossible to achieve a negative value. In this instance, the values have to be whole numbers as all the dots are whole numbers, and by adding whole numbers together it is impossible to achieve the result of a decimal or a fraction that cannot be simplified. Additionally, all the values of n have not been tested, which means that we cannot be certain that it works for all the infinite values.

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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