- Level: International Baccalaureate
- Subject: Maths
- Word count: 1525
Lacsaps fractions are an arrangement of numbers that are symmetrically repeating based on a constant pattern.
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Introduction
Söderportgymnasiet Name: Ali Thaer Abdulrasak
IB Math SL Internal assessment Date: 2012-10-04
Type I
Lacsap’s fractions
In this task the goal is to consider the fractions that are presented in this symmetrically repeating sequence, and to determine a general equation or statement for this pattern. The pattern is as such.
Figure 1 Lacsap’s fraction as given in the assignment
Lacsap’s fractions are an arrangement of numbers that are symmetrically repeating based on a constant pattern. Hence it would make it possible to derive a general statement for this pattern.
Firstly I am asked to find the sixth row in this pattern, to do that I have to identify the difference between each row’s numerators and denominators. For ease of presentation I will call the numerators N and the denominators D, I will eliminate the sides as they are only ones (at least in the beginning) I will also develop a general formula for the numerators and denominators separately.
Numerator:
This is the pattern I have observed for the numerators.
The difference increases, between the numerators of each row, with one. I will construct a table to show this clearlty and to show the numerators for rows 6, 7 and 8.
Table 1 the row number and the denominator values for that row, where n is the row number and N is the denominator
n | N |
1 | 1 |
2 | 3 |
3 | 6 |
4 | 10 |
5 | 15 |
6 | 21 |
7 | 28 |
8 | 36 |
Difference
+2
+3
+4
+5
+6
+7
+8
As you can see the pattern can then be expressed as
Middle
+7
+6
+5
+4
+3
+2
+1
+0
7
28
22
18
16
16
18
22
28
+8
+7
+6
+5
+4
+3
+2
+1
+0
8
36
29
24
21
20
21
24
30
36
Now using the values from tables 1 and 5 I can find the 6th and 7th row.
Now onto finding the general statement for the denominators, I will use the same procedure by equating D as the denominator, n as the row number and r as the element number. It is worth noting that the first element in each row is r=0.
Table 6 the values of the denominator (D), numerator (N) and the row number (n) for the first element
n | N | r | D |
1 | 1 | 1 | 1 |
2 | 3 | 1 | 2 |
3 | 6 | 1 | 4 |
4 | 10 | 1 | 7 |
5 | 15 | 1 | 11 |
It appears to be that the denominator is equal to the difference between the numerator and (n-1) of that row. In other words,
as 1 is the element number
Table 7 the value of the denominator (D) for the first element using the derived equation.
N | N | r | D | |
1 | 1 | 1 | 1 | |
2 | 3 | 1 | 2 | |
3 | 6 | 1 | 4 | |
4 | 10 | 1 | 7 | |
5 | 15 | 1 | 11 |
This equation seems to be correct, but just to make sure I will try it with the second and third elements as well.
Table 8 the value of the denominator for r=2 using the derived equation.
N | N | r | D | |
2 | 3 | 2 | 3 | |
3 | 6 | 2 | 4 | |
4 | 10 | 2 | 6 | |
5 | 15 | 2 | 9 |
The results of the denominator, calculated using the derived equation, do not match the ones in Lacsap’s fractions for the second element. But the formula does seems to work for the 2nd term of the 2nd element but not the rest. I will let the difference be x, hence the equation (for the difference between n and r) should be
Table 9 the value difference between the numerators (N) and the difference between x to see if that would result in D
n | N | r | D | ||
2 | 3 | 2 | 3 | ||
3 | 6 | 2 | 4 | ||
4 | 10 | 2 | 6 | ||
5 | 15 | 2 | 9 |
From the table above I observed that x is too small, hence when subtracted with N it will not yield the same result as the denominator. Hence I will make and equation where h is a number required to be multiplied with x to make it large enough, so that when it is subtracted from N it will equal D.
I will use D=4 and 6. x= 1. N= 6 the previous value seemed to work with the unmodified expression. The rest I will present in a table.
Table 10 showing the values of x, r, h and D for the 2nd element
n | N | r | D | |||
3 | 6 | 2 | 2 | 4 | ||
4 | 10 | 2 | 2 | 6 | ||
5 | 15 | 2 | 2 | 9 |
Conclusion
n r
0
1
2
3
4
5
1
1
1
2
3
2
3
3
6
4
4
6
4
10
7
6
7
10
5
15
11
9
9
11
15
I can see that the values for the denominators match that in figure one hence I can say that this form is correct.
I also know that:
Table 12 the first ten terms of Lacsap’s fractions, to validate the general statement.
n r | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
1 | |||||||||||
2 | |||||||||||
3 | |||||||||||
4 | |||||||||||
5 | |||||||||||
6 | |||||||||||
7 | |||||||||||
8 | |||||||||||
9 | |||||||||||
10 |
Although this general statement seems to work there are still some limitations, for instance n must be bigger the zero, if n is 0 then the denominator of an element maybe undefined. As well as the fact that there cannot be zero rows as then the pattern will be non-existent. Furthermore n has to be positive, as if n was to be negative then substituted into the equation then the answer would be positive even though it is supposed to be negative.
n=-3
Table 13 the numerator values for the negative values of n
n | N |
-3 | 3 |
-4 | 6 |
-5 | 10 |
-6 | 15 |
-7 | 21 |
-8 | 28 |
As you can see a negative n value yielded positive results, not to mention the fact that there can never be any negative rows. r must also be greater than zero as there can never be any negative elements or zero elements as zero elements would imply no pattern. n and r have to also be integers as there can never be any half rows or half elements i.e. the values of n and r cannot be fractions (or decimals).
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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