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Stellar Numbers. In this task geometric shapes which lead to special numbers will be considered.

Extracts from this document...

Introduction

Maths Internal Assessment

Type 1 – Mathematical Investigation

Mathematics Standard Level                                          

Stellar Numbers

September 2011

Luís Ferreira

Teacher: Mr.Robson

Aim – In this task geometric shapes which lead to special numbers will be considered.

For example the easiest of these are square numbers which can be represented by squares of side 1, 2, 3 and 4.

  1. The following diagrams show a triangular pattern of evenly spaced dots. The numbers of dots in each diagram are examples of triangular numbers. Complete the triangular numbers sequence with three more terms. Find a general statement that represents the nth triangular number in terms of n.[1]

image01.pngimage02.png



Finding a general statement:

Now that three more terms have been drawn, the general statement can be found for the sequence; 1, 3, 6, 10, 15, 21, 28, 36.

To do this, the constant difference in the sequence will need to be found, as shown below. This is needed to determine the type of equation (linear, quadratic, cubic etc...) [2]

Number of term (n)

0

1

2

3

4

5

6

7

Sequence

1

3

6

10

15

21

28

36

First Difference

 2              3               4              5               6              7            8                image00.pngimage00.pngimage00.pngimage00.pngimage00.pngimage00.pngimage00.png

Second Difference

        1        1        1                      1                 1           1                    image00.pngimage00.pngimage00.pngimage00.pngimage00.pngimage00.pngimage00.png

The standard rules to find the general statement were researched and the following the method was put into practice for all the shapes in this portfolio.

If the second difference is a constant, the formula for the nth term contains n2 as in a quadratic formula i.e. ax2 + bx + c.

The value of ‘a’ is half the constant difference. In this example a= image18.png

Hence, that the first part of the formula is image18.png n2. To find the rest of the formula, the differences between the values in the sequence and the values of image18.pngn2 will need to be calculated.

Number of term (n)

0

1

2

3

4

5

6

7

Sequence

1

3

6

10

15

21

28

36

image18.pngn2

0

0.5

2

4.5

8

12.5

18

24.5

Difference between Sequence and

image18.png n2

1

2.5

4

5.5

7

8.5

10

11.5

Second difference

image03.pngimage03.pngimage03.pngimage03.pngimage03.pngimage03.pngimage03.pngimage00.pngimage00.pngimage00.pngimage00.pngimage00.pngimage00.pngimage00.png

This second difference illustrates the value for ‘b’ which is equal toimage03.png.

However the value of ‘c’ has not yet been determined. It was calculated using an example:

Using n=2:

image18.png n2+ image03.png n + c = 6
image18.png (2)2+ image03.png (2) + c = 6
5 + c=6
c=1

...read more.

Middle

Now that I have more terms and now that they have been drawn I can now find a general statement for the sequence, using the previous methods.

Again the second difference is the constant therefore the formula for the nth term contains n2 as in the quadratic equation: ax2 + bx + c

The value of ‘a’ is half the constant difference. In this example a= image13.png 
Now that I know that the first part of the formula is
image13.pngn2 I can proceed to find the values of ‘b’ and ‘c’.

pSn

5S0

5S1

5S2

5S3

5S4

5S5

5S6

Sequence

1

6

16

31

51

76

106

image13.pngn2

0

2.5

10

22.5

40

62.5

90

Difference between Sequence and image13.pngn2

1

3.5

6

8.5

11

13.5

16

Second difference

image13.pngimage13.pngimage32.pngimage13.pngimage13.pngimage00.pngimage00.pngimage00.pngimage00.pngimage00.pngimage00.png


This second difference tells me the value for ‘b’ which is equal to
image13.png. To find the value of ‘c’ I will use the previous methods:

Using n=4:

image13.png n2+image13.png n + c = 51
image13.png (4)2+image13.png (4) + c = 51
50 + c = 51
c = 1

To check that these are the correct values, two more examples were used:

Using n=1:

image13.png n2+image13.png n + 1 = 6
image13.png (1)2+image13.png (1) + 1 = 6

Using n=6:

image13.png n2+image13.png n + 1 = 106
image13.png (6)2+image13.png (6) + 1 = 106

Therefore the general statement for this shape is:

image13.pngn2 +image13.pngn + 1

As this is a quadratic equation, a graph was plotted to demonstrate how it expanded

image33.png

Now, a p value of 6 will be used based on a regular hexagon:

image34.png

image35.png

Stage Number

Number of Dots

Notes and observations

6S0

1

None

6S1

7

Adding 6 to previous

6S2

19

Adding 6x2  to previous

6S3

37

Adding 6x3  to previous

6S4

61

Adding 6x4  to previous

6S5

91

Adding 6x5  to previous

6S6

127

Adding 6x6  to previous

Again, another pattern related to the pattern in other shapes has been detected. The relationship discovered is that the number of dots in the next stage is equal to the number of dots in the previous stage plus the term of the current stage multiplied by 6.  

I.e.:  6Sn = 6Sn-1+ 6n

To prove this equation an existing example from above will be used.
For Stage 2:
6Sn = 6Sn-1 + 6n
6S2= 6S1 + (6x2)
=
7+ 12
=19


For Stage 4:
6Sn = 6Sn-1 + 6n
6S4= 6S3 + (6x4)
=
37+ 24
=61

Finding a general statement:

Now that I have more terms and now that they have been drawn I can now find a general statement for the sequence, using the previous methods.

pSn

6S0

6S1

6S2

6S3

6S4

6S5

6S6

Sequence

1

7

19

37

61

91

127

First Difference

           6            12             18             24             30           36                           image00.pngimage00.pngimage00.pngimage00.pngimage00.pngimage00.png

Second Difference

                  6               6               6               6                6                               image00.pngimage00.pngimage00.pngimage00.pngimage00.pngimage00.png

Again the second difference is the constant therefore the formula for the nth term contains n2 as in the quadratic equation: ax2 + bx + c

The value of ‘a’ is half the constant difference. In this example a= image36.png = 3

Now that I know that the first part of the formula is 3n2 I can proceed to find the values of ‘b’ and ‘c’.

pSn

6S0

6S1

6S2

6S3

6S4

6S5

6S6

Sequence

1

7

19

37

61

91

127

3n2

0

3

12

27

48

75

108

Difference between Sequence and n2

1

4

7

10

13

16

19

Second difference

               3            3               3               3                  3              3          image00.pngimage00.pngimage00.pngimage00.pngimage00.pngimage00.png

This second difference tells me the value for ‘b’ which is equal to 3. To find the value of ‘c’ I will use the previous methods:

Using n=4

3n2+3n + c = 61
3(4)
2+3(4) + c = 61
60 + c = 61
c = 1

To check that these are the correct values, two more examples were used:

Using n=6:

3n2+3n + 1 = 127
3(6)
2+3(6) + 1 = 127

Using n=2:

3n2+3n + c = 19
3(2)
2+3(2) + 1 = 19

Therefore the general statement for this shape is:

3n2 +3n + 1

As this is a quadratic equation, a graph was plotted to demonstrate how it expanded:
image37.png

Finally, a p value of 8 will be analysed:image38.png

image39.png


Once again, another pattern was observed as the stage numbers developed. This time, the connection discovered is that the number of dots in the next stage is equal to the number of dots in the previous stage plus the term of the current stage multiplied by 16.  

Stage Number

Number of Dots

Notes and observations

8S0

1

None

8S1

17

Adding 16 to previous

8S2

49

Adding 16x2  to previous

8S3

97

Adding 16x3  to previous

8S4

161

Adding 16x4  to previous

8S5

241

Adding 16x5  to previous

8S6

337

Adding 16x6  to previous

I.e.:  8Sn = 8Sn-1+ 16n

For Stage 3:
8Sn = 8Sn-1 + 16n
8S3= 6S2 + (16x3)
=
49+ 48
=97

For Stage 6:
8Sn = 8Sn-1 + 16n
8S6= 8S5 + (16x6)
=
241+ 96
=337

Finding a general statement:

Now that I have more terms and now that they have been drawn I can now find a general statement for the sequence, using the previous methods.

pSn

8S0

8S1

8S2

8S3

8S4

8S5

8S6

Sequence

1

17

49

97

161

241

337

First Difference

         16            32             48             64             80           96                           image00.pngimage00.pngimage00.pngimage00.pngimage00.pngimage00.png

Second Difference

                  16             16            16             16              16                               image00.pngimage00.pngimage00.pngimage00.pngimage00.pngimage00.png

Again the second difference is the constant therefore the formula for the nth term contains n2 as in the quadratic equation: ax2 + bx + c

The value of ‘a’ is half the constant difference. In this example a= image40.png = 8

Now that I know that the first part of the formula is 8n2 I can proceed to find the values of ‘b’ and ‘c’.

pSn

8S0

8S1

8S2

8S3

8S4

8S5

8S6

Sequence

1

17

49

97

161

241

337

8n2

0

8

32

72

128

200

288

Difference between Sequence and n2

1

9

17

25

33

41

49

Second difference

               8            8               8               8                  8               8         image00.pngimage00.pngimage00.pngimage00.pngimage00.pngimage00.png

This second difference tells me the value for ‘b’ which is equal to 8. To find the value of ‘c’ I will use the previous methods:

Using n=4

8n2+8n + c = 161
8(4)
2+8(4) + c = 161
160 + c = 161
c = 1

To check that these are the correct values, two more examples were used:

Using n=2:

8n2+8n + 1 = 49
8(4)
2+8(4) + 1 = 49

Using n=3:

8n2+8n + 1 = 97
8(3)
2+8(3) + 1 = 97

Therefore the general statement for this shape is:

8n2 +8n + 1

As this is a quadratic equation, a graph was plotted to demonstrate how it expanded:

image41.png

  1. Hence, produce the general statement, in terms p and n, that generates the sequence of p-stellar numbers for any value of p at stage Sn.

Polygon

pSn

a

b

c

1

3Sn

image42.png

image43.png

1

2

6Sn

6

6

1

3

4Sn

4

4

1

4

5Sn

image44.png

image44.png

1

5

6Sn

3

3

1

6

8Sn

8

8

1

When analyzing all the examples given, it is clear that there is no regular general pattern for the values of a, b and c that relate to the values of p and n. For polygons 2, 3 and 6 it seems that the values of a and b are the equal to the values of p. However, for polygons 4 and 5 it appears the values of a and b are half the value of p.

Nevertheless, I noticed that some shapes (such as polygon 2, 3 and 6) can be considered to have double their number of vertices, if we include the points that go in i.e. the concave lines. This arouses the question: what are vertices? A vertex should be considered “the common endpoint of two or more rays or line segments (…) Vertex typically means a corner or a point where lines meet.” [3] If we took this and followed it exactly then polygon 2 would have 12 vertices, instead of 6, polygon 3 would have 8, instead of 4, and polygon 6 would have 16 vertices, instead of 8. Using this these terms I can now find a general statement:

Polygon

pSn

a

b

c

General Statement

1

3Sn

image45.png

image43.png

1

...read more.

Conclusion

Accepting that all points at which lines meet as vertices is also important for the matter, as a stellar shape is assumed to be one that has the same number of concave and convex points. Another limitation is that the values in the sequence, the values of p and the values of n all have to be positive, as a negative value in each of these would be impossible to draw.

  1. Explain how you arrived at the general statement.

Initially I attempted to see whether there was a relationship between the number of dots and the values of ‘a’ and ‘b’. Seen there was none, I decided to create a table so that the ideas were more organized. I then, started to see the connection.

At first I did not consider the concave points in some of the polygons to be vertices; if we were strictly speaking they are p-pointed stars. However when I considered these to be vertices and altered my general table I could see the immediate relationship between the values of ‘a’ and ‘b’ and the value of p. Since ‘a’ was equal to ‘b’ in most cases it was simple to find the relationship.

As a final point c was continuously 1 therefore I believe it to be 1 for the general expression derived.           


[1]All the questions in dark blue are from the Oporto British School Maths internal Assessment handout 2011

[2] Steps followed by “The nth term of quadratics” at http://www.pearsonpublishing.co.uk/education/samples/S_492153.pdf

[3]. Definition from http://www.mathopenref.com/vertex.html

...read more.

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