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LACSAP's Functions

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Introduction

LACSAP’S FRACTIONS

Tanya Zhandria Waqanika
Prince Andrew High School
IB Math SL Portfolio #1
Mr Brown

Task 1 – Find the Numerator
Task 2 – Plot relation between the row number and Numerator, write general statement about it
Task 3 – Find the 6
th and 7th rows
Task 4 – Find the general statement for E
n(r)
Task 5 – Find additional rows with General Statement
Task 6 – Discuss the scope/limitations of the General Statement

Task 1 - Finding the Numerator

Since ‘LACSAP’ is just ‘Pascal’ reversed, I decided to do some research about Pascal’s triangle. Through my research, I was able to come across the equation nCr, or n!/r!(n –r)! where n represents the row number and r represents the ‘element’ number, or the diagonal row number.

e.g. 3C2 = 3!/2!(3 – 2)! = 1 x 2 x 3/1 x 2 x 1 = 3

 This equation is used to find any number within Pascal’s triangle. With 1 at the top of both pyramids representing n = 0, r = 0, I decided to compare Pascal’s triangle to LACSAP’s fractions.
                                           
image00.jpg
                                                              Pascal’s triangle

1

1

1

image01.png

1

1

image08.png

image08.png

1

1

image10.png

image11.png

image10.png

1

1

image02.png

image12.png

image12.png

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1


                                                                  LACSAP’S Fractions.

Since I’m trying to find the numerator, I was looking for any similarities between the two triangles.

...read more.

Middle

th an 7th Numerators

6th Numerator
(6+1)C2
(7)
C2
6!/2!(6 – 2)!
6!/2!4! =
21

7th Numerator

(7+1)C2
(8)
C2
8!/2!(8– 2)!
8!/2!6! =
28

These numbers can also be found going down r = 2 of Pascal’s triangle.

Finding the Denominator

Whilst it took a bit of trial of error, I found that the difference between the Numerator and the Denominator were following a pattern similar to Pascal’s Triangle, eg. In this grid, I have replaced the numbers with the differences between the Numerator and Denominator to clearly see the pattern.

0

0

1

1

1

1

image04.png

image04.png

1

1

image05.png

image06.png

image05.png

1

1

image06.png

image07.png

image07.png

image06.png

1

image09.png

Comparing the new triangle to Pascal’s Triangle, it almost appears that the new triangle is a mirror image of Pascal’s triangle. Using this new information, I decided to plot the row numbers of LACSAP’s triangle against r = 1 of the new triangle as it’s the only diagonal row that appears consistent with Pascal’s triangle.

n (Row number)

Difference of Numerator and Denominator

1

0

2

1

3

2

4

3

5

4

According to the table, the difference of the Numerator and Denominator is (n -1) as the values for the differences, though the same, are only the same once they’ve been bumped down one.

...read more.

Conclusion

n(r), can be written as:

Numerator/Denominator = (n+1)C2/( (n+1)C2) - r(n - r)


Task 5 – Find Additional Rows

Row 8
(8+1)C2/( (8+1)C2) - r(8 - r)

Element Number

(n+1)C2/( (n+1)C2) - r(n - r)

0

(8+1)C2/( (8+1)C2) - 0(8 - 0) = 1

1

(8+1)C2/( (8+1)C2) - 1(8 - 1) = 36/29

2

(8+1)C2/( (8+1)C2) - 2(8 - 2) = 36/24

3

(8+1)C2/( (8+1)C2) - 3(8 - 3) = 36/21

4

(8+1)C2/( (8+1)C2) - 4(8 - 4) = 36/20

After this, the rest of the row is reflected, thus it looks like
1, 36/29, 36/24, 36/21, 36/20, 36/21, 36/24, 36/29, 1

Row 9
(9+1)C2/( (9+1)C2) - r(9 - r)

Element Number

(n+1)C2/( (n+1)C2) - r(n - r)

0

(9+1)C2/( (9+1)C2) - 0(9 - 0) = 1

1

(9+1)C2/( (9+1)C2) - 1(9 - 1) = 45/37

2

(9+1)C2/( (9+1)C2) - 2(9 - 2) = 45/31

3

(9+1)C2/( (9+1)C2) - 3(9 - 3) = 45/27

4

(9+1)C2/( (9+1)C2) - 4(9 - 4) = 45/25

1, 55/47, 55/41, 55/37, 55/35, 55/35, 55/37, 55/41, 55/47, 1

Row 10
(10+1)C2/( (10+1)C2) - r(10 - r)

Element Number

(n+1)C2/( (n+1)C2) - r(n - r)

0

(10+1)C2/( (10+1)C2) - 0(10 - 0) = 1

1

(10+1)C2/( (10+1)C2) - 1(10 - 1) = 55/46

2

(10+1)C2/( (10+1)C2) - 2(10 - 2) = 55/39

3

(10+1)C2/( (10+1)C2) - 3(10 - 3) = 55/34

4

(10+1)C2/( (10+1)C2) - 4(10 - 4) = 55/31

5

(10+1)C2/( (10+1)C2) - 5(10 - 5) = 55/20

1, 55/46, 55/39, 55/34, 55/31, 55/20, 55/31, 55/34, 55/39, 55/46, 1

Task 6 - Discuss the scope/limitations of the General Statement

Some limitations of General Statement are that:

  • The numerator must be greater than 0
  • In the equation (n+1)C2,  n +1 must be greater than 2/r in order for the General Statement to work

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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