Lacsap's Fractions

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  • Topic: Number, Fraction, Mathematics in medieval Islam
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  • Published : November 25, 2012
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Lacsap’s Fractions
IB Math 20 Portfolio
By: Lorenzo Ravani

Lacsap’s Fractions Lacsap is backward for Pascal. If we use Pascal’s triangle we can identify patterns in Lacsap’s fractions. The goal of this portfolio is to find an equation that describes the pattern presented in Lacsap’s fraction. This equation must determine the numerator and the denominator for every row possible.

Numerator
Elements of the Pascal’s triangle form multiple horizontal rows (n) and diagonal rows (r). The elements of the first diagonal row (r = 1) are a linear function of the row number n. For every other row, each element is a parabolic function of n. Where r represents the element number and n represents the row number. The row numbers that represents the same sets of numbers as the numerators in Lacsap’s triangle, are the second row (r = 2) and the seventh row (r = 7). These rows are respectively the third element in the triangle, and equal to each other because the triangle is symmetrical. In this portfolio we will formulate an equation for only these two rows to find Lacsap’s pattern. The equation for the numerator of the second and seventh row can be represented by the equation: (1/2)n * (n+1) = Nn (r) When n represents the row number. And Nn(r) represents the numerator Therefore the numerator of the sixth row is Nn(r) = (1/2)n * (n+1) Nn(r) = (1/2)6 * (6+1) Nn(r) = (3) * (7) Nn(r) = 21 Figure 2: Lacsap’s fractions. The numbers that are underlined are the numerators. Which are the same as the elements in the second and seventh row of Pascal’s triangle.

Figure 1: Pascal’s triangle. The circled sets of numbers are the same as the numerators in Lacsap’s fractions.

Graphical Representation
The plot of the pattern represents the relationship between numerator and row number. The graph goes up to the ninth row. The rows are represented on the x-axis, and the numerator on the y-axis. The plot forms a parabolic curve, representing an exponential increase of the numerator compared to the row number. Let Nn be the numerator of the interior fraction of the nth row. The graph takes the shape of a parabola. The graph is parabolical and the equation is in the form: Nn = an2 + bn + c The parabola passes through the points (0,0) (1,1) and (5,15) At (0,0): 0 = 0 + 0 + c ! ! At (1,1): 1 = a + b ! ! ! At (5,15): 15 = 25a + 5b ! ! ! 15 = 25a + 5(1 – a) ! 15 = 25a + 5 – 5a ! 15 = 20a + 5 ! 10 = 20a! ! ! ! ! ! ! ! therefore c = 0 therefore b = 1 – a Check with other row numbers At (2,3): 3 = (1/2)n * (n+1) ! (1/2)(2) * (2+1) ! (1) * (3) ! N3 = (3)

therefore a = (1/2) Hence b = (1/2) as well

The equation for this graph therefore is Nn = (1/2)n2 + (1/2)n ! which simplifies into !

Nn = (1/2)n * (n+1)

Denominator
The difference between the numerator and the denominator of the same fraction will be the difference between the denominator of the current fraction and the previous fraction. Ex. If you take (6/4) the difference is 2. Therefore the difference between the previous denominator of (3/2) and (6/4) is 2.

!

!

Figure 3: Lacsap’s fractions showing differences between denominators

Therefore the general statement for finding the denominator of the (r+1)th element in the nth row is:

Dn (r) = (1/2)n * (n+1) – r ( n – r )
Where n represents the row number, r represents the the element number and Dn (r) represents the denominator. Let us use the formula we have obtained to find the interior fractions in the 6th row.

Finding the 6th row
- First denominator ! ! ! ! ! ! ! ! ! ! ! ! - Second denominator ! ! ! ! ! ! ! ! ! ! ! ! ! denominator = 6 ( 6/2 + 1/2 ) – 1 ( 6 – 1 ) ! = 6 ( 3.5 ) – 1 ( 5 ) ! = 21 – 5 = 16 denominator = 6 ( 6/2 + 1/2 ) – 2 ( 6 – 2 ) ! = 6 ( 3.5 ) – 2 ( 4 ) ! = 21 – 8 = 13

! !

-Third denominator ! ! ! ! ! ! ! ! ! ! ! ! - Fourth denominator ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! - Fifth denominator ! ! ! ! ! ! ! ! ! ! ! ! ! ! !

! ! ! ! !

denominator = 6 ( 6/2 + 1/2 ) – 3 ( 6 – 3 ) ! = 6 ( 3.5 ) – 3 ( 3 ) !...
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