• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  17. 17
    17
  18. 18
    18
  19. 19
    19
  20. 20
    20

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.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Maths Internal Assessment -triangular and stellar numbers

    In other words the general statement for the 6-stellar numbers is (12 x (n - 1)) + 1. However, this expression is not complete as it does not apply to the 6-stellar numbers. Therefore, as previously stated, for the triangular numbers, n-1 = n(n-1)/2.

  2. Stellar numbers

    value added to the next term is +1 greater than the value of the term before. Hence from this the next 3 triangle stages may be derived. In the 6th, 7th and 8th stages, the number of dots become respectively: 21, 28, 36.

  1. Stellar Numbers Portfolio. The simplest example of these is square numbers, but over the ...

    S1=1 S2=3 S3=6 S4=10 S5=15 S6=21 S7=28 S8=36 S9=45 S10=55 The first pattern that I noticed when looking at this information was that the order looks like this when broken down: 1, 1+2, 1+2+3, 1+2+3+4, 1+2+3+4+5, ... , 1+2+3+4+5+6+7+8+9+10 The first attempt I made was saying the general term was simply (n+1).

  2. Comparing the surface area of different shapes with the same volume

    Utilization of GDC (Graphic Display Calculator) Y= (tab) Y1= x2+4*x* (1000/x2) Y2= ?*x2+2*?*x*(1000/(?*x2)) TBLSET (Table Set) Window Xmin=0 Xmax=20 Xscl=10 Ymin=0 Ymax=1000 Yscl=100 Xres=1 Then to calculate the minimum of both curve lines, the following steps must be followed: 2nd-calc-3: minimum Left bound-right bound-Guess *The same steps must be taken for the second line also Cylinder Prism 6.

  1. Stellar Numbers Portfolio. In this task I will consider geometric shapes, which lead ...

    So, in this case =16n because of the multiples of 16, the ( to because each of the numbers multiplied by 16 is one less than the of the term, and add 1 because the addition of terms starts initially from 1.

  2. Stellar Numbers. In this task geometric shapes which lead to special numbers ...

    If I add 12x6 to the previous number (253) I can calculate the number of dots in stage 7. 6S7 = 253 + (12x6) = 337 A relationship was established in the sequence. This was determined by observing the sides of the shape and analyzing the table.

  1. Stellar Numbers. In this study, we analyze geometrical shapes, which lead to special numbers. ...

    of n provides exactly the same results as the consecutive formula from Task 1. Finally, one can confirm that the two formulas are identical by the following calculation: Substituting with the general statement: and (n-1) Opening the brackets, one can see that the two sides are identical; therefore the general

  2. Stellar numbers and triangular numbers. Find an expression for the 6-stellar number at ...

    You can see that at each stage there are six vertices on the shape and so we can say that each shape has a "6-stellar number". This is the total number of dots in the above diagram and there is a way in which we can find the number of dots.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work