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

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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.

- 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:

- 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 | ...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. - 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 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
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