# Ib Math Sl Ia Lacsap's Fractions

Topics: Fraction, Mathematics in medieval Islam, Number Pages: 6 (1396 words) Published: January 22, 2013
In Lacsap’s Fractions, En(r) refers to the (r+1)th term in the nth row. The numerator and denominator are found separately, therefore to find the general statement, two different equations, one for the numerator and one for the denominator, must be found. Let M=numerator and let D=denominator so that En(r) = M/D.

To find the numerator for any number of Lacsap’s Fractions, an equation must be made that uses the row number to find the numerator. Because the numerator changes depending on the row, the two variables (row number and numerator) must be compared. To find this equation, the relationship between the row number and numerator must be found, put it graph form, and the equation for the graph will be the equation needed.

Row Number, n| Numerator, N|
1| 1|
2| 3|
3| 6|
4| 10|
5| 15|

Numerator
Numerator
Row Number
Row Number

The equation for the numerator can be derived by using quadratic regression on a graphing calculator. The equation is; y = .5x2 + .5x. This translates into; M=.5n2+.5n, where n=row number, and M=numerator. This means that any numerator from a certain row number can be found by using this equation. For example, to find the numerator of the sixth row, “6” needs to be substituted in for n. M= .5n2 + .5n

M= .5(6)2 + .5(6)
M= .5(36) + .5(6)
M= 18 + 3
M= 21
The Numerator for row six is 21
They method to find the equation for the denominator is similar, but slightly more difficult because there is more than one variable changing for each number of Lacsap’s fractions. All of the numbers in a row had the same numerator, but the denominator changes depending on which row the number is, and which term in the row it is.

Example: 1/1, 10/7, 10/6, 10/7, 1/1
In this example all of the numerators are 10 (1/1 can also be written as 10/10), but the denominators change.
So, to find the equation of the denominator, a relation must be set up between the denominator, D, and the term number, r, instead of the Numerator and the row number. However, because each row is unique, each row will have its own quadratic, therefore, a relation must be set up for each row. Ex graph and data table:

Term #| D|
0| 10|
1| 7|
2| 6|
3| 7|
4| 10|

After the graphs are made, the equations should be found by using quadratic regression on a graphing calculator.

The equations turn out to be:
Row 1: x2 - x + 1
Row 2: x2 - 2x + 3
Row 3: x2 - 3x + 6
Row 4: x2 - 4x + 10
Row 5: x2 - 5x + 15

If these five equations are analyzed, it is plain to see that in ax2 + bx + c form, a=1, b=-n, and c=M. So that n= the row number and M=the numerator. Once r is substituted in for x, the equation for the denominator is complete: D = r2 – nr + (.5n2+.5n).

Now that both parts of the general statement are found, the two can simply be substituted in for M and D. En(r)=M/D
En(r)=(.5n2+.5n)/r2-nr+(.5n2+.5n)
General Statement:
(.5n2 + .5n)
r2 – nr + (.5n2 + .5n)
(.5n2 + .5n)
r2 – nr + (.5n2 + .5n)

En(r) =

Now that the general statement for En(r) has been found, it is just a matter of substitution to find any row. Row 6
En(r)=(.5n2+.5n)/r2-nr+(.5n2+.5n)
E6(0) = (.5(6)2 + .5(6)) / ((0)2 – (6)(0) + (.5(6)2 + .5(6)) E6(0) = (.5(36) + 3) / (.5(36) + 3)
E6(0) = (18 + 3) / (18 + 3)
E6(0) = 21/ 21
E6(0) = 1
En(r)=(.5n2+.5n)/r2-nr+(.5n2+.5n)
E6(1) = (.5(6)2 + .5(6)) / ((1)2 – (6)(1) + (.5(6)2 + .5(6)) E6(1) = (.5(36) + 3) / (1 – 6 + (.5(36) + 3))
E6(1) = (18 + 3) / (-5 + (18 + 3))
E6(1) = 21/ (-5 + 21)
E6(1) = 21/ 16

En(r)=(.5n2+.5n)/r2-nr+(.5n2+.5n)
E6(2) = (.5(6)2 + .5(6)) / ((2)2 – (6)(2) + (.5(6)2 + .5(6)) E6(2) = (.5(36) + 3) / (4 – 12 + (.5(36) + 3))
E6(2) = (18 + 3) / (-8 + (18 + 3))
E6(2) = 21/ (-8 + 21)
E6(2) = 21/ 13

En(r)=(.5n2+.5n)/r2-nr+(.5n2+.5n)
E6(3) = (.5(6)2 + .5(6)) / ((3)2 – (6)(3) + (.5(6)2 + .5(6)) E6(3) = (.5(36) + 3) / (9 – 18 + (.5(36)...