LACSAP’S Fractions - Math SL Type I Name: Yao Cia Hua Date ... · applied. By using pattern 1, the value 6 can simply be added to the current numerator value, which is 15. This

Yao Cia Hua Mathematics SL LACSAP’S Fraction-‐ Portfolio Type I

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LACSAP’S Fractions - Math SL Type I

Name: Yao Cia Hua

Date: March 22nd, 2012

Teacher: Mr. Mark Bethune

School: Sinarmas World Academy


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Lacsap triangle is a reversed Pascal triangle. This task focuses mainly on finding the relationship between the number of row n and the numerator N and also the relationship between the element of a row r and the denominator D . Through this, a general statement base for En (r) onN andD are suppose to be stated and explained. For this task various technologies, such as Geogebra, MathType and calculator, are needed in order to produce a more organized piece of work and clearer graphs and diagrams.

Finding The Numerator In The Sixth Row:

In order to find the sixth row of this of this triangle, a pattern must firstly be found.

As it is seen from the diagram on the above, the pattern shows that, by adding 1 to each of the difference between the 2 previous numerators, this will equate the numerator of the next row. For example: (using the first 2 numerators) Difference: 3 – 1 = 2 The following numerator: 2 + 1 = 3


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There is also another way of looking at these patterns. If the difference between the 2 previous numerators is already known, the following numerator can be found by summing the difference and the value of the current numerator, finally adding 1 . For example: (using the first 2 numerators) Difference: 3 -1= 2 Current numerator: 3 The following numerator: (2 + 3) + 1 = 6 So, in order to find the numerator of the sixth row, either pattern 1 or 2 can be applied. By using pattern 1 , the value 6 can simply be added to the current numerator value, which is 15 . This will then give an answer of 21 . By using the second pattern, the value 5 , which is the difference between 15 and 10 , can be added to 15 then add 1 to find the value. Either way will result the same value. The Relationship & Finding A General Statement For The Numerator (N):


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When only looking at these points above, the relationship between the number of row,n , and the value of the numerator, N , without drawing any curve, it seems to be exponential. However, when the graph is continued as shown above, it can be identified that all the co-ordinate points lay of the same parabola. A parabola has a general function of y = Ax2 + Bx +C . As there are 3 variables in this function, A , BandC , 3 functions are required. Process Of Finding A General Statement For Numerator (N ):

(1;1),(2;3),(3;6)1= (A × (12 ))+ (B ×1)+C1= A + B +C3= (A × (22 ))+ (B × 2)+C3= 4A + 2B +C6 = (A × (32 ))+ (B × 3)+C6 = 9A + 3B +C

1= A + B +C.........13= 4A + 2B +C........26 = 9A + 3B +C.........32 −1:2 = 3A + B........43− 2 : 3= 5A + B...........55 − 4 :1= 2A

A = 12

sub.A = 12

1= 12+ B +C.........1'

3= (4 × 12)+ 2B +C.......2 '

12= B +C.........1''

3= 2 + 2B +C..........2 ''1= 2B +C..........2 '''

2 '''−1'' : 12= B

B = 12

sub.A = 12& B = 1

2

1= 12+ 12+C..........1'''

1= 1+CC = 0

Sub. A = 12,B = 1

2,C = 0 to y = Ax2 + Bx +C

y = 12x2 + 1

2x

y = 12x(x +1)


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Replace x & ywith row ( n ) & Numerator (N ):

N = 12n × (n +1)

N = 12(n2 + n)

N = n2 + n2

N = n(n +1)2

General statement for numerator (N ):

Nn =n(n +1)2

Finding The Sixth & Seventh Rows: Since the general statement for the numerator is already found in the previous question, in order to find the numerators of both the sixth and seventh rows, values will only need to be substituted in to the general statement.

N6 =6(6 +1)2

= 6 × 72

= 21

N7 =7(7 +1)2

= 7 × 82

= 28

The Pattern, The Relationship & Finding A General Statement For The Denominator (D ):

In order to find the value of denominator for the 6th and 7th rows, a pattern from the given data must firstly be discovered.


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It is shown from the diagram above that by adding 1 to each of the difference between the 2 previous denominators, it gives the value of the next denominator in the next row. This means, in order to find the denominator of both the 6 th and 7 th rows,1 must be added to the previous difference depending what the difference is for each element. For example: using the 2 nd & 3 rd rows: Difference: 4 – 2 = 2 The denominator of the next row ( 4 th row) = (difference +1 ) + current denominator = (2 +1 )+ 4 = 7 Since, the numerator for both the 6 th and 7 th rows are known, base on the pattern for the value of denominator, row 6 and row 7 is also findable.

⇒ 6 th row: 1, 2116, 2113, 2112, 2113, 2116,1

⇒ 7 th row: 1, 2822, 2818, 2816, 2816, 2818, 2822,1


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However, a general statement for its denominator D is also required to find the general statement for the different fractions, called elements, in a particular row.

Points of curve A Points of curve B


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Proofing That These Points Create Parabolas:

Curve A Curve B These 2 images above prove that the shapes of the curves of both functions are supposed to be parabolic. This is so because, in order to make the difference between the D values equal, the differences between them have to be calculated twice. The branches represent “the difference between them is…”. In order to find the difference, the first value must be subtracted by the following value. For example: Difference between the first 2 values in the table on the right: 11− 9 = 2


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Process Of Finding A General Statement For The Denominator (D): Using the 5 th row: (1;11),(2;9),(3;9)y = Ax2 + Bx +C

11= (A × (12 ))+ (B ×1)+C.......19 = (A × (22 ))+ (B × 2)+C......29 = (A × (32 ))+ (B × 3)+C.......311= A + B +C...........1'9 = 4A + 2B +C........2 '9 = 9A + 3B +C.........3'2 '−1' :−2 = 3A + B......43'− 2' :0 = 5A + B........55 − 4 :2 = 2AA = 1

sub.A = 10 = (5 ×1)+ B......5 '0 = 5 + BB = −5sub.A = 1& B = −59 = (4 ×1)+ (2 × (−5))+C......2 ''9 = 4 + (−10)+C9 = −6 +CC = 15

Sub. A = 1 ,B = −5 ,C = 15 to actual function y = Ax2 + Bx +C :

y = (1× x2 )+ ((−5)× x)+15y = x2 − 5x +15

Replacing x & ywith r & D :

D = r2 − 5r +15 Using the 4 th row: (1;7),(2;6),(3;7)y = Ax2 + Bx +C

7 = (A × (12 ))+ (B ×1)+C.......16 = (A × (22 ))+ (B × 2)+C......27 = (A × (32 ))+ (B × 3)+C.......37 = A + B +C...........1'6 = 4A + 2B +C........2 '7 = 9A + 3B +C.........3'2 '−1' :−1= 3A + B......43'− 2' :1= 5A + B........55 − 4 :2 = 2AA = 1

sub.A = 11= (5 ×1)+ B......5 '1= 5 + BB = −4sub.A = 1& B = −46 = (4 ×1)+ (2 × (−4))+C......2 ''6 = 4 + (−8)+C6 = −4 +CC = 10


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Sub. A = 1 ,B = −4 ,C = 10 to actual function y = Ax2 + Bx +C

y = (1× x2 )+ ((−4)× x)+10y = x2 − 4x +10

Replacing x & ywith r & D :

Explanation: Base on the 2 equations found for both the 5 th and 4 th rows, when compared with the lacsap triangle, it appears that the value of c represents the numerator of the current row, whereas the value of b represents the difference between the current numerator and the previous numerator. General statement for the denominator (D ):

Dn = r2 − ((Nn − Nn−1)× r)+ Nn

Finding A General Statement for En (r) : Here, r represents the number of element in a row and n represents the row number. As moving to the left of a row, the element of that row will be denoted as

r +1,r + 2,r + 3... . For example: E2 (1) =32

SinceEn (r) will give the value of an element, including both the numerator, N , and

the denominator, D , in a row, it can be concluded thatEn (r) =Nn

Dn

.

En (r) =

n(n +1)2

r2 − ((Nn − Nn−1)× r)+ N

However, since the general equation forN has already been found above, N in the

denominator can be substituted by n(n +1)2

, so the whole equation will only consist

of variables n and r .

D = r2 − 4r +10


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En (r) =

n(n +1)2

r2 − ((Nn − Nn−1)× r)+ Nn

En (r) =

n(n +1)2

r2 − ((n(n +1)2

− ((n −1)(n −1+1))2

)× r)+ n(n +1)2

En (r) =

n(n +1)2

r2 − (((n(n +1))− ((n −1)n)2

)× r)+ n(n +1)2

En (r) =

n(n +1)2

r2 − ((n2 + n − n2 + n

2)× r)+ n(n +1)

2

En (r) =

n(n +1)2

r2 − ((2n2)× r)+ n(n +1)

2

En (r) =

n(n +1)2

r2 − nr + n(n +1)2

General statement for En (r) :

En (r) =

n(n +1)2

r2 − nr + n(n +1)2


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Testing the validity of the general statement En (r) :

In order to know whether this general statement is working properly and works for every value in the lacsap triangle, different row number, n , and different number of elements, r , will be substituted in to the equation. 1st Test: 2nd Test:

n = 3,r = 1

E3(1) =

3(3+1)2

(12 )− (3×1)+ (3+1)32

E3(1) =

122

1− 3+ 6

E3(1) =64

n = 7,r = 4

E7 (4) =

7(7 +1)2

(42 )− (7 × 4)+ 7(7 +1)2

E7 (4) =

562

16 − 28 + 562

E7 (4) =2816

By doing the 2 tests above, the results prove that the general equation is valid and gives the right value when a specific row number and element number are substituted in to the equation. Finding additional rows to check validity: 8th row: E8 (1),E8 (2),E8 (3),E8 (4),E8 (5),E8 (6),E8 (7)

E8 (1) =

8(8 +1)2

(12 )− (8 ×1)+ 8(8 +1)2

= 3629

E8 (2) =

8(8 +1)2

(22 )− (8 × 2)+ 8(8 +1)2

= 3624

E8 (3) =

8(8 +1)2

(32 )− (8 × 3)+ 8(8 +1)2

= 3621


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E8 (4) =

8(8 +1)2

(42 )− (8 × 4)+ 8(8 +1)2

= 3620

E8 (5) =

8(8 +1)2

(52 )− (8 × 5)+ 8(8 +1)2

= 3621

E8 (6) =

8(8 +1)2

(62 )− (8 × 6)+ 8(8 +1)2

= 3624

E8 (7) =

8(8 +1)2

(72 )− (8 × 7)+ 8(8 +1)2

= 3629

⇒ 8th row: 1 , 3629

, 3624

, 3621

, 3620

, 3621

, 3624

, 3629

, 1

9th row: E9 (1),E9 (2),E9 (3),E9 (4),E9 (5),E9 (6),E9 (7),E9 (8)

E9 (1) =

9(9 +1)2

(12 )− (9 ×1)+ 9(9 +1)2

= 4537

E9 (2) =

9(9 +1)2

(22 )− (9 × 2)+ 9(9 +1)2

= 4531

E9 (3) =

9(9 +1)2

(32 )− (9 × 3)+ 9(9 +1)2

= 4527

E9 (4) =

9(9 +1)2

(42 )− (9 × 4)+ 9(9 +1)2

= 4525

E9 (5) =

9(9 +1)2

(52 )− (9 × 5)+ 9(9 +1)2

= 4525


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E9 (6) =

9(9 +1)2

(62 )− (9 × 6)+ 9(9 +1)2

= 4527

E9 (7) =

9(9 +1)2

(72 )− (9 × 7)+ 9(9 +1)2

= 4531

E9 (8) =

9(9 +1)2

(82 )− (9 × 8)+ 9(9 +1)2

= 4537

⇒ 9th row: 1 , 4537

, 4531

, 4527

, 4525

, 4525

, 4527

, 4531

, 4537

, 1

Scope & Limitation Of The General Statement En (r) :

Types of Values

Value Proof Validity

Positive integers

excluding 0

n > 0 , r > 0

Valid

Equals to 0

n = 0 , r = 0

n = 0,r = 0

E0 (0) =

0(0 +1)2

(02 )− (0 × 0)+ 0(0 +1)2

E0 (0) =

0202

=∅

Invalid

n = 3,r = 1

E3(1) =

3(3+1)2

(12 )− (3×1)+ (3+1)32

E3(1) =

122

1− 3+ 6

E3(1) =64


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n > 0 , r = 0

n = 3,r = 0

E3(0) =

3(3+1)2

(02 )− (3× 0)+ 3(3+1)2

E3(0) =

122

0 − 0 + 122

E3(0) =

122122

E3(0) =122× 212

E3(0) = 1

Valid

Negative integers

excluding 0

n < 0 , r < 0

n = −2,r = −5

E−2 (−5) =

−2(−2 +1)2

(−5)2 + ((−2)× (−5))+ −2(−2 +1)2

E−2 (−5) =

(−2)× (−1)2

25 +10 + (−2)× (−1)2

E−2 (−5) =135

Invalid

Irrational number

n = 2 , r = π

n = 2,r = π

E 2 (π ) =

2( 2 +1)2

π 2 − ( 2 ×π )+ 2( 2 +1)2

E 2 (π ) =

2 + 22

π 2 −π 2 + 2 + 22

Invalid


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By using the table above, it can be concluded that the general statement is only valid for r -values that are equal to 0 and greater than 0. r can be 0 because this variable is not a part of the numerator in the general statement. On the other hand, the general equation is only valid when the n -values is greater than 0. 0 is excluded because when 0 is substituted, the numerator will always equal to 0. This makes the whole fraction undefined. However, for both cases, positive values that are substituted will exclude fractions and irrational numbers. Justification Of The General Statement: When finding a general statement for the denominator, which named D , there were some difficulties because it has to consist both n , the row number, and also r , the number of element in a row. Since n and r couldn’t be plotted against each other,

must be plotted against . At first, it was difficult to see the pattern in the 2 equations for the denominator calculated above. However, when looking at the 2 equations, the value that is substituted for C appear to represent the numerator in the current row. For example: 5th row: Function: Current numerator: 15⇒C Previous numerator: 10 Difference between current and previous numerator: 15 −10 = 5⇒ B So, when the values are replaced by another notation, dominatorD is equal to: Dn = r

2 − ((Nn − Nn−1)× r)+ Nn .

As mentioned before, that En (r) =Nn

Dn

, so to find a general statement for it, the

general statement for both the numerator and denominator must be substituted. However, since there are also Nn is also a part of the general equation for denominator, it must be replaced by the general statement of the numerator leaving it with the 2 required variables, r and n . Finally, after all the substitutions and the

calculations, the general statement for En (r) appears to be

n(n +1)2

r2 − nr + n(n +1)2

.

r D

D = r2 − 5r +15


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Work Cited: http://www.thestudentroom.co.uk/showthread.php?t=1865309

http://www.fluther.com/131327/can-anyone-help-me-with-lascaps-fractions/

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LACSAP’S Fractions - Math SL Type I Name: Yao Cia Hua Date ... · applied. By using pattern 1, the value 6 can simply be added to the current numerator value, which is 15. This