5th row with the numerator value 15
y =0.5(5)2 + 0.5(5)
y =0.5(25) + 2.5
y =12.5 + 2.5
y =15
Why don’t we try to use this formula to find the next several rows to see how accurate and reliable it really is?
According to the triangular numbers, row 6 should have a numerator of 21.
y =0.5(6)2 + 0.5(6)
y =0.5(36) + 3
y =18 + 3
y =21
According to the triangular numbers, row 7 should have a numerator of 28.
y=0.5(7)2 + 0.5(7)
y=0.5(49) + 3.5
y=24.5 + 3.5
y= 28
According to the triangular numbers, row 8 should have the numerator of 36.
y=0.5(8)2 + 0.5(8)
y=0.5(64) + 4
y=32 + 4
y= 36
Table 1: The Row Number of the Lacsap’s Fraction and the corresponding numerator
In order to find out which equation best fits the data, the coefficient of determination will be used. Coefficient of determination (R2) is used to deduce how well future outcomes can be predicted using the model which represents the data set. A data model with an R2 value of 1 is the best predictor of future trends. As the value of R2 value of a model gets closer to 1, it becomes a better representative of the data trends. Hence the function with the closest R2 value will be the best model of the data.
Graph 1: Row Number VS Numerator with a Linear Trendline
Graph 2: Row Number VS Numerator with a Linear Trendline
Out of the 2 functions used to represent the data, quadratic function is a better model. In fact, it is safe to state that quadratic function is the best model of the data as the R2 value is exactly 1. This means no other model can be a better representative of this data set. Thus the quadratic function will be used as the data model for this data set.
The equation of the line of best fit is 0.5x2+0.5x+(4x10-14). This equation is similar to the one derived using algebra; the difference being the 4x10-14. The constant at the end does not make a sizeable impact on the equation as its magnitude, 10-14, renders it almost insignificant. Thus the graphing method complies with the equation found using the algebra; hence it can be concluded that the equation for the numerator is correct.
Now, let us look at the trend for the denominator. The given triangle is symmetrical, meaning, the numbers on the left side have identical and matching numbers on the right side, thus it is more like a mirror image.
However, let us look at the denominators in each row and how they vary according to the element number. Let us take the 6th to the 8th row where the difference is more blatant.
6th row
7th row
8th row
What can be noticed from evaluating the 6th to the 8th row are that for every odd row (7th row), the difference between the elements and their denominators is an even value. And for every even numbered row, the difference between the elements and their denominators is an odd value. In addition, in each row, there is a parabolic pattern. For instance, in the 8th row, the difference between the first two elements is 5, and then it reduces in an odd numbered sequence till it reaches the minimum of 1. Once it touches 1, it increases the same way up.
By looking at the graph, we can safely deduce that the rows are definitely parabolic. But what is the significance of having them parabolic?
Let us look at the 3 equations that we obtain:
6th row: x2 - 6x + 21
7th row: x2 – 7x + 28
8th row: x2 – 8x + 36
What we notice from here is that the variable of ‘b’ in each other rows is the row number (n) and the variable of ‘c’ that we obtain is the numerator for each of the given rows and we have ‘x’ which is the element number.
Let us generalize these 3 equations to make one equation:
y = x2- (n)x + numerator
y = numerator + x2 - (n)x
y = numerator + x (x - n)
y = numerator - x( n - x)
numerator - r(x - r)where ‘r’ is the element number and ‘x’ is the row number.
How about we test this formula for several rows
Row 5 has a numerator of 15.
1st element should have the denominator 11 and the 2nd element should have a denominator of 9. (Excluding the first and last elements of the row valued at ‘1’)
1st element
=15 - 1(5-1)
=15 – 4
=11
2nd element
=15 - 2(5 - 2)
= 15- 6
= 9
Row 6 has a numerator of 21.
1st element
=21-1(6-1)
=21-5
=16
2nd element
=21-2(6-2)
=21-8
= 13
3rd element
=21-3(6-3)
=21-9
=12
4th element
=21-4(6-4)
=21-8
= 13
5th element
=21-5(6-5)
=21 – 5
=16
It is important to keep in mind the symmetry that had been mentioned previously. The 1st and 2nd element mirror the 4th and the 5th elements in terms of its denominator.
We have now obtained the formulas for both the numerator and for the denominator according to its element. Let us compile both the elements together and then proceed to test its accuracy and validity.
→ Numerator = 0.5x2 + 0.5x, where x is the row number.
→ Denominator = numerator - r(x - r)where ‘r’ is the element number and ‘x’ is the row number.
This makes it: 0.5x2 + 0.5x – r(x -r)
By using the formula above to acquire the numerators and the denominators for the 7th, 8th, 9th and 10th row, we will be able to check the extent of accuracy of the formulas.
7th row
Numerator
y=0.5(7)2 + 0.5(7)
y=0.5(49) + 3.5
y=24.5 + 3.5
y= 28
Denominator (should have a total of 8 elements with ‘1’ as the first and the last)
1st element
=28-1(7-1)
=28-6
=22
2nd element
=28-2(7-2)
=28-10
=18
3rd element
=28-3(7-3)
=28-12
= 16
4th element
=28-4(7-4)
=28-12
=16
5th element
=28-5(7-5)
=28-10
=18
6th element
=28-6(7-6)
=28-6
=22
Therefore, the 7th row will look like this:
Let us try the same thing on the 8th row
Numerator
y=0.5(8)2 + 0.5(8)
y=0.5(64) + 4
y=32 + 4
y= 36
Denominator (should have a total of 9 elements where ‘1’ is the first and last)
1st element
=36-1(8-1)
=36-7
=29
2nd element
=36-2(8-2)
=36-12
=24
3rd element
=36-3(8-3)
=36-15
=21
4th element
=36-4(8-4)
=36-16
=20
5th element
=36-5(8-5)
=36-15
=21
6th element
=36-6(8-6)
=36-12
=24
7th element
=36-7(8-7)
=36-7
=29
Therefore, the 8th row will look like this:
Let us continue with the 9th row
Numerator:
=0.5(9)2 + 0.5(9)
=0.5(81) + 4.5
=40.5 + 4.5
=45
Denominator (should have 10 elements with ‘1’ as the first and the last)
1st element
=45-1(9-1)
=45-8
=37
2nd element
=45-2(9-2)
=45-14
=31
3rd element
=45-3(9-3)
=45-18
=27
4th element
=45-4(9-4)
=45-20
=25
5th element
=45-5(9-5)
=45-20
=25
6th element
=45-6(9-6)
=45-18
=27
7th element
=45-7(9-7)
=45-14
=31
8th element
=45-8(9-8)
=45-8
=37
Therefore, the 9th row will look like this:
Now if we combine both the equations of the numerator and the denominator to immediately find out the fraction, we can obtain a general formula.
→ Numerator = 0.5x2 + 0.5x, where x is the row number.
→ Denominator = Numerator – r(x-r) where ‘r’ is the element number and ‘x’ is the row number.
This makes it: 0.5x2 + 0.5x – r(x -r)
NUMERATOR
DENOMINATOR
Thus, we can also write it as:
→ 0.5x2 + 0.5x
0.5x2 + 0.5x – r(x -r)
Let us try this equation to solve the 9th and 10th rows
The 9th row has 10 elements (with ‘1’ as the first and the last). It should also have a common numerator of 45.
Element 1
= 0.5(9) 2 + 0.5(9)
(0.5(9) 2 + 0.5(9)) – 1(9-1)
= 45
37
Element 2
= 0.5(9) 2 + 0.5(9)
(0.5(9) 2 + 0.5(9)) – 2(9-2)
= 45
31
Element 3
= 0.5(9) 2 + 0.5(9)
(0.5(9) 2 + 0.5(9)) – 3(9-3)
= 45
27
Element 4
= 0.5(9) 2 + 0.5(9)
(0.5(9) 2 + 0.5(9)) – 4(9-4)
= 45
25
Element 5
= 0.5(9) 2 + 0.5(9)
(0.5(9) 2 + 0.5(9)) – 5(9-5)
= 45
25
Element 6
= 0.5(9) 2 + 0.5(9)
(0.5(9) 2 + 0.5(9)) – 6(9-6)
= 45
27
Element 7
= 0.5(9) 2 + 0.5(9)
(0.5(9) 2 + 0.5(9)) – 7(9-7)
= 45
31
Element 8
= 0.5(9) 2 + 0.5(9)
(0.5(9) 2 + 0.5(9)) – 8(9-8)
= 45
38
The 10th row should have 11 elements (with ‘1’ as the first and the last).
Element 1
= 0.5(10) 2 + 0.5(10)
(0.5(10) 2 + 0.5(10)) – 1(10-1)
= 55
46
Element 2
= 0.5(10) 2 + 0.5(10)
(0.5(10) 2 + 0.5(10)) – 2(10-2)
= 55
39
Element 3
= 0.5(10) 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) 2 + 0.5(10)
(0.5(10) 2 + 0.5(10)) – 6(10-6)
= 55
31
Element 7
= 0.5(10) 2 + 0.5(10)
(0.5(10) 2 + 0.5(10)) – 7(10-7)
= 55
34
Element 8
= 0.5(10) 2 + 0.5(10)
(0.5(10) 2 + 0.5(10)) – 8(10-8)
= 55
39
Element 9
= 0.5(10) 2 + 0.5(10)
(0.5(10) 2 + 0.5(10)) – 9(10-9)
= 55
46
We know that the numerator for the 10th row is 55 because based on Pascal’s triangular numbers; the 10th row has a value of 55 in its 3rd element (from the left). And as for the denominator, we are certain of it because symmetry can be observed through a median of numbers from the first element to the 9th element (excluding ‘1’ from the first and the last) out of 11 elements.
Conclusion
The general statement:
→ 0.5x2 + 0.5x
0.5x2 + 0.5x – r(x -r)
However, the concluding general statement has various limitations. Firstly, the formula is unable to calculate the value of ‘1’ that begins and ends every row. Secondly, the value of ‘x’ has to be greater than or equal to 1. The value of ‘x’ also cannot be a negative number as any negative number when squared will equal to a positive value. Although 0 rows does not make any practical sense, it can also be considered a mathematical limitation nonetheless.
Firstly, when scanning the given data, a pattern can be observed between the numerators and denominators as the row progresses. By formulating a statement for the numerator and the denominator based on the patterns, I managed to produce a general statement that would allow me to find the rest of the fractions of any row that I wish because it is understood that because there is a pattern, there is a consistency.
This general statement can be considered reliable as it as been proven accurate through thorough evaluation of various numbers and instances. Using Pascal’s triangular numbers and the given graph for the numerator and the observation of symmetry and constant evaluation and experimentation for the denominator, we can conclude confidently that the given general statement is reliable.