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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]

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 Second Difference 1        1        1                      1                 1           1

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=

Hence, that the first part of the formula is  n2. To find the rest of the formula, the differences between the values in the sequence and the values of n2 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 n2 0 0.5 2 4.5 8 12.5 18 24.5 Difference between Sequence and  n2 1 2.5 4 5.5 7 8.5 10 11.5 Second difference

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

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

Using n=2:

n2+  n + c = 6
(2)2+  (2) + c = 6
5 + c=6
c=1

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=
Now that I know that the first part of the formula is
n2 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 n2 0 2.5 10 22.5 40 62.5 90 Difference between Sequence and n2 1 3.5 6 8.5 11 13.5 16 Second difference

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

Using n=4:

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

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

Using n=1:

n2+ n + 1 = 6
(1)2+ (1) + 1 = 6

Using n=6:

n2+ n + 1 = 106
(6)2+ (6) + 1 = 106

Therefore the general statement for this shape is:

n2 +n + 1

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

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

 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 Second Difference 6               6               6               6                6

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=  = 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

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:

Finally, a p value of 8 will be analysed:

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 Second Difference 16             16            16             16              16

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=  = 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

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:

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 1 2 6Sn 6 6 1 3 4Sn 4 4 1 4 5Sn 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

1

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

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