• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

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 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

image02.png

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.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Extended Essay- Math

    � ( � ( � ( � ( � �^����C��?��3/4���\ � �u��!?��no�/? �_j���@P@P@P@P@P@P@P@P@P@P@/�w����Å�p����(r)P �E� �:�����l?�7�-������ ��� � ( � ( � ( � ( � ( � ( � ( � ( � ( � ( �?-��;�P���8o��t�(������O�6Û����_�W�...ptP@P@P@P@P@P@P@P@P@P@P@���(p�g��7���:k"�Qt����d'����e��"�B�?�:(1/2����]]�4�ko=�"ú�JÈW� �...2I`�]�r��ß��QOٷ�o�t��z��1/4�kR|6� ��&'i�jV>3��(r)k~�3/4-�-~"�͡x�G�6�4=n W�+����51/2{M����;- ��1/4J��h�-1/4 y�(tm)�3/4�t����Cd$�|0�-+�Xd3/4"3/4�� �!��%��5Y1/4G�E�[]I8P�����>���x��j_4��_|��k�(r)�º"���^o�w�1/2�'��-�[�]�u�o�Ì*�(c)"i"@��$�sz��&k� �w/��wH�1/4á-�+Z ��1/4W�%"M�^B&"R��T���(tm)b�-Z�(c)��I��o��'�o���k�/��;�> �� x.�O�5�c�>�eÕµ[�0-'iVp�� ...��*(c)�Z}�� k�"u��s�1�1/4?f� x

  2. Salida del sol en NY

    vez mas sim�trica, por lo tanto tender�n a igualarse lo que causa que el d�a durara, lo mismo que la noche, entonces si se quiere que hacer un modelo se tendr� que tomar que tan cerca del ecuador estamos, as� poder saber que entre m�s cerca del la hora del alba ser� aproximadamente semejante al d�a anterior.

  1. Population trends. The aim of this investigation is to find out more about different ...

    curvature that isn't very much alike what the real situation of other countries has been like. According to the research made in order to introduce the investigation the many restrictions imposed on the population has meant that this model is a good one but only for China because there is a way to keep an order in the population growth.

  2. Maths Project. Statistical Analysis of GCSE results at my secondary school summer 2010 ...

    28 f 62 46 St 9 40 46 m 86 45 Sw 9 40 46 m 86 44 Sz 10 40 46 m 86 43 Te 10 40 40 f 80 42 Th 10 40 40 m 80 41 Ti 11 40 40 f 80 40 To 10 40 34

  1. Investigating Sin Functions

    Therefore, the graph changes by a factor of 1/|b|. We can further see this by taking a look at multiple graphs on one window: Black = (y = sin(x)), Red = (y = sin(10x)), Blue is sin(5x); As we can see from the graph, as B increases, the graph compresses, or shrinks more.

  2. Derivative of Sine Functions

    �repeat its values every 2,(so the function is periodic with a period of 2) �the maximum value is a, the minimum value is -a �the mean value of the function is zero �the amplitude of the function is a According to the characteristics the form of gradient function is similar to a cosine function with parameter of a.

  1. MATH IB SL INT ASS1 - Pascal's Triangle

    = r² - nr + (0.5n² + 0.5n) This is the explicit formula for the denominator. To test the validity of the statement, I calculated the denominators of the 5th up to the 9th row with the general formula, to compare the data with the table above (Fig.

  2. Lacsap triangle investigation.

    2 + 0.5(10) (0.5(10) 2 + 0.5(10)) ? 3(10-3) = 55 34 Element 4 = 0.5(10) 2 + 0.5(10) (0.5(10) 2 + 0.5(10)) ? 4(10-4) = 55 31 Element 5 = 0.5(10) 2 + 0.5(10) (0.5(10) 2 + 0.5(10)) ? 5(10-5) = 55 30 Element 6 = 0.5(10)

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work