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# LACSAP's Functions

Extracts from this document...

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 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. Pascal’s triangle

 1 1 1 1 1  1 1   1 1    1

LACSAP’S Fractions.

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

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  1 1   1 1    1 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.

Conclusion

n(r), can be written as:

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

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

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

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