Lacsap's Fractions

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  • Topic: Mathematics in medieval Islam, Number, Quadratic equation
  • Pages : 8 (1727 words )
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  • Published : February 28, 2013
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Type I – Mathematical Investigation
Lacsap’s Fractions

The focus of this investigation is surrounding Lascap’s Fractions. They are a group of numbers set up in a certain pattern. A similar mathematical example to Lacsap’s Fractions is Pascal’s Triangle. Pascal’s Triangle represents the coefficients of the binomial expansion of quadratic equations. It is arranged in such a way that the number underneath the two numbers above it, is the sum. Ex. 1

1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

In the example of Pascal’s Triangle below, the highlighted numbers represent this pattern. The two numbers above the third add up to equal the third. (e.g. 2+1=3)

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

Although Pascal’s Triangle is set up similarly to Lacsap’s Fractions, the patterns are different and the numbers are set up in fractions as opposed to whole numbers. The first five rows of Lacsap’s Fractions are set up like this:

1

1 3/2 1

1 6/4 6/4 1

1 10/7 10/6 10/7 1

1 15/11 15/9 15/9 15/11 1

In the next part of the investigation, we will explore the patterns of Lacsap’s Fractions and how to find the following rows. Part 1:

To begin the investigation, we are given the first five rows of Lacsap’s Fractions as shown below.

1 1

1 3/2 1

1 6/4 6/4 1

1 10/7 10/6 10/7 1

1 15/11 15/7 15/7 15/11 1

To figure out the various patterns, it would be easiest to split the fractions into two parts. Let’s start with including only the numerators in the same pyramid pattern. Notice that in each row, the numerator is constant throughout. Now, if we separate the fractions and create the pyramid with only the numerators, we come to the set of numbers below.

1 1

1 3 1

1 6 6 1

1 10 10 10 1

1 15 15 15 15 1

If we focus on the highlighted numbers, we can see a pattern forming regarding the numerators. The differences between each number and the next number increase each time by one. Hence:
3 – 1= 2
6 – 3= 3
10 – 6= 4
15 – 10= 5

Also, notice how the difference of the numerators equals the row number of the larger numerator. This can be condensed into saying that R2 – R1= n, when R2 is the larger numerator, R1 is the smaller numerator, and n is the row number.

These data can be represented in a table showing the primary and secondary differences of the numerators as shown below.

Row #12345
Numerator1361015
Primary Difference23
45
Secondary Difference111

This table not only shows the first differences between the numerators, but also the second differences. Because the second differences of all the numerators are equal, this means that this is a geometric series.

Part 2:

In this next part of the investigation, we will apply the patterns we found in the previous part to determine the numerator of the next row.

One way that we can find the numerator of the next row would be to use the pattern of the differences between the numerators. We can continue the pattern that we began in the previous part.

3 – 1= 2
6 – 3= 3
10 – 6= 4
15 – 10= 5

If we use this pattern, we assume that the difference between the next numerator and the numerator of row 5 will be 6. So we substitute the variable x in the equation x – 15= 6. To solve for x, we need to add 15 to 6 to isolate the x. The answer we come up with should be 21. We can check this by doing 21 – 15= 6. So our numerator is 21.

But if we do not have the previous rows and we want to still...
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