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  • Level: GCSE
  • Subject: Maths
  • Word count: 3363

In this investigation I will explore the relationship between a series of straight, non-parallel, infinite lines on a plane surface and analyze the number of lines, maximum number of crossover points and open and closed regions.

Extracts from this document...

Introduction

Aim: In this investigation I will explore the relationship between a series of straight, non-parallel, infinite lines on a plane surface and analyze the number of lines, maximum number of crossover points and open and closed regions. I will investigate patterns that emerge from the collected data (relating to number of lines, the maximum number of crossover points and the maximum number of open and closed regions obtained).

Method: I will use diagrams to illustrate my investigation, and use mathematical notation in the form of tables to describe the sequences that appear and apply what I have learnt about sequences to determine formulas or ‘rules’ to predict the results for more lines.

In the course work hand out we were given, we were presented with a diagram which had four lines, five cross-over points and a total of ten regions. For the purposes of my investigation I will start with 1 line and tabulate my findings (with regard to number of lines, the maximum number of crossover points and the maximum number of open and closed regions). I will redraw the diagrams adding one more line every time until I have a diagram with six lines. I should then have enough information to be able to predict the results for a diagram with 7 lines with the use of the formulae, which I will find to summarise the rules for the sequences.

Diagram 1: 1 Line (open regions are depicted with numbers)

image00.png

In this diagram there are:

Number of Lines (n)

Cross-Over Points

Open Regions

Closed Regions

Total Regions

1

0

2

0

2

Diagram 2A

...read more.

Middle

4

6

8

10

12

First Difference

+2

+2

+2

+2

+2

I found that the values in the ‘First Difference’ line are the same or ‘constant’ number  + 2. The terms are even numbers or multiples of 2. For each new (non-parallel) line that is added, there will be two more Open Regions. I can summarise this rule for working out even numbers with a formula like this:

Thus, OR(n) = 2(n) by inspection

I can check my formula by using data I have already collected and know to be correct.

OR (2) using the formula 2(n) let n = 2

2(2) = 4                             CORRECT!

 OR (3) using the formula 2(n) let n = 3

2(3) = 6                             CORRECT!

OR (5) using the formula 2(n) let n = 5

2(5) = 10                        CORRECT!

I predict:

OR (7) using the formula 2(n) let n = 7

2(7)  = 14

To test the prediction I made I draw a diagram with 7 lines. Counting the open regions confirms the result of 14

image02.png

This diagram, where lines (n)  = 7 and every line crosses every other line,  (no lines are parallel) proves that the formula I used to predict the amount of Open Regions

2(n) is correct because there are 14 Open Regions in this diagram, and that is what I predicted.

Step 2:

I will list the number of cross-over points to investigate the pattern in the second sequence.

The number of Cross –Over Points is given by:         COP(n)

(n) is the number of lines (where n can be any natural number) then

By inspection,          COP(1) = 0

                        COP(2) = 1

                        COP(3) = 3

                        COP(4) = 6

                        COP(5) = 10

                        COP(6) = 15

This sequence can be written as:

Terms (n)

1

2

3

4

5

6

Value Cross-Over Points

0

1

3

6

10

15

First Difference

+1

+2

+3

+4

+5

Second Difference

+1

+1

+1

+1

Since the values of the first difference are not the same, I find the second difference. I found that the values in the ‘Second Difference’ line are the same or ‘constant’. From what I have learnt in class, I know that the polynomial for this sequence of values is quadratic (an expression is quadratic if the highest power of the variable is 2). It is of the form

2

an   + bn + c = 0

Investigation (i)

To solve for the sequence I will find (b) where:

(n) can be defined as the number of lines or terms

(a) can be defined as ½ the second difference

(c) can be defined as the zero term where the n term is equal to ‘0’

let         n = 1

        a = ½

        c = -1

(½) x (1)2 + b x (1) – 1 = 0        

½ + b - 1 = 0

b = ½

Substitute (b) as found above into the formula to check its accuracy for the first term in the sequence COP(1).

         b = ½

        n = 1

        a = ½

        c = -1

(½) x (1)2 + (½) x (1) – 1  = COP(1)

(½) + (½) –1 = COP(1)

0 = COP(1) – which is correct

Investigation (ii)

To solve for the sequence I will find (b) where:

(n) is the number of lines or terms

(a)is ½ the second difference

(c)is the zero term where the n term is equal to ‘0’

let         n = 2

        a = ½

        c = -1

(½) x (2)2 + b x (2) – 1 = 0        

(½) x (4)+ b x (2) - 1 = 0

2b = -1

b = -½

Substitute (b) into the formula to check its accuracy for the second term in the sequence COP(2).

(½) x (2)2 + (-½) x (2) – 1  = COP(2)

(½) x (4) + (-½) x (2)  –1 = COP(2)

2 + (-1) -1 = COP(2)

0 = COP(2) – which is not correct

It would seem that the formula:  an2   + bn + c = 0 does not work for the sequence COP(n). However I have noticed another possibility if I were to rearrange the formula using the values for (a), (b) and (c) that I found in Investigation (i)

As before let:

b = ½(as found in Investigation (i))

a = ½ (can be defined as ½ the second difference)

c = -1 (can be defined as the zero term where the n term is equal to ‘0’)

COP (n)        = ½ n2 + ½ n –1

                  =  n2        + n  - 1

                = ½n (n+1) – 1

                = n(n+1)/2 – 1

COP (2) is the second term of the sequence let n = 2        

COP (2) = 2 (2 +2)/2 – 1

         =8/2 – 1

        = 4 – 1

        = 3

COP (2)  is not 3. Therefore this formula does not work.

Investigation (iii)

I will investigate a new formula for the sequence of numbers.        

If I start the sequence with 1. Add 2 to get 3. Add 3 to get 6. Add 4 to get 10. Continue the pattern to get: 1, 3, 6, 10, 15, 21, 28, 36, 45, …

I notice that these numbers are triangular numbers.
The terms of this sequence or ‘triangular numbers’ increase in a way that can be represented by triangular patterns of dots like this:

        COP(2)        COP(3)       COP(4)      COP(5)             COP(6)

image03.png

            1                      3                 6                  10                15

The formula for producing them is: n(n + 1)/2

I will apply this formula to the sequence COP (n)

[where (n) is the number of lines in the diagram]

COP(2) let n = 2

2 (2+1)

=       2

=     6
       2

=    3

COP (3) let n = 3

3 (3+1)

=       2

=     12
       2

=    6

I can check that the formula is true for triangular numbers by looking at the dot pattern diagrams above.

But the sequence for Cross-over Points is:

COP(1) = 0

COP(2) = 1

COP(3) = 3

COP(4) = 6

COP(5) = 10

COP(6) = 15

COP (2) is not equal to 3 and COP (3) is not equal to 6. Therefore this formula does not work.

Investigation (iiii)

The same problem keeps emerging for all the formulas for Cross-over Points I have tried so far. However I can see that the sequence for COP (n) and Triangular numbers are very similar. The only difference is this:

The first term of the sequence COP (1) is: 0.

The first term of a sequence of triangular numbers is: 1.

When I looked at Diagrams 1- 6 again it became clear that when no lines are parallel, each new line intersected exactly once with each previous line. Thus, when the nth line is added, it makes  (n-1) new intersections or cross-over points.

I decided to incorporate this concept into the formula I have for discovering the nth term in a sequence of triangular numbers. So instead of  

n (n+1)n (n-1)

         2                I’ll try                  2                [where (n) is the number of lines in the diagram]

To test this formula I will substitute some known terms into the equation.

COP(2) let n = 2

2 (2-1)

=       2

=     2
       2

=    1                CORRECT!

COP (3) let n = 3

3 (3-1)

=       2

=     6
       2

=    3                CORRECT!        

COP (6) let n = 6

6 (6-1)

=       2

=     30
       2

=    15          CORRECT!

I predict:

COP(7) using the formula n (n-1)

                                        2                

[where (n) is the number of lines in the diagram]  

let n = 7

= 7 (7-1)

       2

= 42

    2

= 21

...read more.

Conclusion

TR(n)

(n) is the number of lines in the diagram (where ‘n’ can be any natural number) then

By inspection,          TR(1) = 2

                        TR(2) = 4

                        TR(3) = 7

                        TR(4) = 11

                        TR(5) = 16

                        TR(6) = 22

From looking at the information I gathered when looking at Diagram 6 I noticed the following:

Number of Lines (n)

Cross-Over Points

Open Regions

Closed Regions

Total Regions

1

0

2

0

2

2

1

4

0

4

3

3

6

1

7

4

6

8

3

11

5

10

10

6

16

6

15

12

10

22

Open Regions + Closed Regions = Total Regions

Based on this observation, I will investigate the following theory:

Adding the two formulas I have found for OR(n) (Open Regions) and CR (n) (Closed Regions) together, should create an accurate formula to define the sequence TR(n)

OR (n) is defined by the formula:         2(n)

CR (n) is defined by the formula:           (n2 –3n + 2)

2

Let TR(n) =2(n) + (n2 –3n + 2)

                              2

To test this formula I will substitute some known terms into the equation. I can use the known sequence for TR (n) to check that the answers I get when using the formula are correct.

(i)TR(1) let n = 1

TR(1)  = 2(1) + (12 –3(1) + 2)

                                         2

TR(1)          = 2 + (1-3 +2)

                          2

TR(1)          = 2         CORRECT!

(ii)TR(2) let n = 2

TR(2)  = 2(2) + (22 –3(2) + 2)

                                         2

TR(2)          = 4 + (4-6 +2)

                          2

TR(2)          = 4         CORRECT!

(iii)TR(3) let n = 3

TR(3)  = 2(3) + (32 –3(3) + 2)

                                         2

TR(3)          = 6 + (9-9 +2)

                         2

TR(3)          = 7         CORRECT!

(iv)TR(6) let n = 6

TR(6)  = 2(6) + (62 –3(6) + 2)

                                         2

TR(6)          = 12 + (36-18 +2)

                          2

TR(6)          = 22         CORRECT!

I predict:

TR(7) using the formula TR(n) = 2(n)+ (n2 –3n + 2)

                                                                    2                

[where (n) is the number of lines in the diagram]  

let n = 7

TR(7)  = 2(7)+  (72 –3(7) + 2)

                                    2

TR(7)  = 14 + 49 –21 + 2

                               2

CR(7)  = 14  + 30

                         2

CR(7)  = 14 + 15

CR(7)  = 29

To test the prediction I made, I draw a diagram with 7 lines.

image02.png

This diagram, where lines  = (n)  = 7 and every line crosses every other line,  (no lines are parallel) proves that the formula I used to predict the amount of total regions in the sequence TR (7) is correct because there are 29 total regions in this diagram, and that is what I predicted.

...read more.

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