Ib Math Portfolio Lacsap's Fractions

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  • Topic: Fraction, Number, Mathematics in medieval Islam
  • Pages : 8 (1884 words )
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  • Published : April 1, 2013
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Exploration of Lacsap’s Fractions

The following will be an investigation of Lacsap’s Fractions, that is, a set of numbers that are presented in a symmetrical pattern. It is an interesting point that ‘Lacsap’ is ‘Pascal’ backwards, which hints that the triangle below will be similar to “Pascal’s Triangle”. 1 1

1 1
1 1
1 1
1 1
There are many patterns evident in this triangle, for instance I can see that there is a vertical axis of symmetry down the middle of the triangle. Each row starts and ends with the number 1. Each row has one more variable than the number of rows, i.e. row 1 has 2 variables. The numerators in the middle stay the same and the diagonals form sequences. In order to decipher the pattern in the numerators and denominators, I had to look at the triangle a different way. Knowing that the numerators of the row don’t change, it occurred to me that the number 1s on the outside of the triangle could be expressed as fractions.

This proves that all the numerators of the row are the same. To further investigate the numerators, I will examine the relationship between the row number and the numerator, which is shown in the table below. These are the numerators after having changed the 1s on the outside of the triangle to their fraction forms, thus making all the numerators the same. Row Number (n)| Numerator in Each Row (N)|

1| 1|
2| 3|
3| 6|
4| 10|
5| 15|

From this table, I can see that the numerator is the result of the current row number added to the numerator of the previous row. That is, if you’re looking for the numerator of row 4, take that number and add the numerator from row 3, which is 6. 4+6=10, and 10 is the numerator of row 4, as shown in the table above. Knowing this, I predict the numerator of the 6th row to be 21. That is, 15 + 6, because 15 was the previous term, and 6 is the current number of rows. In an attempt to prove my prediction, I will formulate an expression for calculating N through Implicit form (otherwise known as recursive). Implicit form is a way to define the terms of a sequence based on previous terms. In other words I can find the numerator of a row based on previous numerators. In this expression, let the numerator be N and the row number be n.

I will now test my expression for its validity using the data from the row numbers and numerators we already know. It is known that the numerator of the 2nd row is 3. This may be used to test the above expression. =+2

= +2
= 1+2
= 3

Hence, it can be concluded that the above expression is likely true. This expression has limitations to using it, such as, it is necessary to know the numerator of the previous row.

Using graphmatica, I will now plot the data shown in the table above showing the row number and the numerator in a graph to better comprehend the relationship between the two.

This graph resembles half of a parabola, which tells me that it is a quadratic function that is opening upwards. Quadratic functions have a ‘U’ shape, and like the triangle, a quadratic function has an axis of symmetry down the middle (symmetric to the y-axis). The general equation of a quadratic function is y= + bx + c. That tells me that there is a squared term in the relationship between the numerator and the row number. It is important to show the relationship between the two so that I can figure out the numerator if given the row number, that is, to show that the numerator is a function of the row number. To derive an expression for calculating the relationship between the two, it is necessary to explore some relationships. Row Number (n)| Row Number| Numerator (N)|

1| 1| 1|
2| 4| 3|
3| 9| 6|
4| 16| 10|
5| 25| 15|

With the aid of the above table, I noticed...
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