Upcoming SlideShare
×

# Math IA

7,317 views

Published on

0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
7,317
On SlideShare
0
From Embeds
0
Number of Embeds
2
Actions
Shares
0
200
0
Likes
0
Embeds 0
No embeds

No notes for slide

### Math IA

1. 1. Michael Pollock IB Math SL Period 7 March 31st 2011Lacsaps Fractions 1 1 3 1 1 2 6 6 1 1 4 4 10 10 10 1 1 7 6 7 15 15 15 15 1 1 11 9 9 11Finding the Numerator EquationThere are many patterns to be found in Lascaps Fractions. In order to meet our first objective offinding the numerator of the sixth row, I only viewed the numerators. This helped to single out patternswhich would eventually lead to an equation. _ _ _ 3 _ _ 6 6 _ _ 10 10 10 _ _ 15 15 15 15 _I took out the sides because they are simply ones, their numerators could be edited to match the rest oftheir respective rows, however their denominators would have to be changed as well.I put the numerators on a chart next to the row number just to get a clearer look at the possible patternsthat could be occurring:Row Number Numerator Value2 33 64 105 15
2. 2. There is a clear pattern here. Firstly, all numerators in a single row are identical. This helps simplify ourequation as we only need the row number and not the position within said row.To get from the numerator in row two, which is three, to the numerator in row three, which is six, weadd three to the numerator of row two. To get from the numerator in row three (6) to row four (10), weadd four, and on it goes. We can observe that the amount we add is increased by one each row. To findthe numerator of row n, we multiply n by half of n plus one, or as follows: n1 N  n=n   2This was derived with a bit of logic and a bit of guess and check. I knew that I would have to multiplyn by some version of itself. To find this value I simply experimented with different values andeventually found one that worked with all the rows I was given. I used this to solve the next row for anumerator of 21, and when I checked by adding 6 to 15 (as in the previously found pattern), I found thesame value.When this equation is expanded we get this: n2 +n N ( n)= 2The expansion is used in order to graph. If we graph this we would be comparing the row number n tothe numerator number N. When we compare these we get this graph depicted below, with the rownumber on the x-axis and the numerator on the y-axis:As the row number increases, the numerator increases in greater increments. We observe exactly this inour triangle. For each subsequent row, the numerator increases by one more than it did for the previousrow. This shows that our equation matches the relationship accurately, even when it comes to highernumbers, without having to test those higher numbers.
3. 3. Using our equations we can easily find the numerators for the sixth and seventh rows:Numerator of Row Six 21Numerator of Row Seven 28And we can check this work easily by using the same pattern we found to begin with. The numerator ofthe fifth row (15) plus six is in fact equal to 21, just as our equation told us. And the numerator of thesixth plus seven is equal to 28, which is what our equation told us row number seven would be.
4. 4. Finding the Denominator EquationNow that we can find the value for the numerator, we need to find the value of the denominator. I willonce again draw the triangle, this time only showing denominators: _ _ _ 2 _ _ 4 4 _ _ 7 6 7 _ _ 11 9 9 11 _In this case we must take into account the row number and the position within the row to find ourvalues. This means that we need to add another variable to our equation.Once again though, there was a pattern that I observed. I will show the numerator and denominator ofthe first fraction in each row, which is located at what well call position one from now on:Row Number Numerator Value Denominator Value2 3 23 6 44 10 75 15 11We can view the difference between the numerator and denominator of each row in position one. Inrow two, they have a difference of one. In row three the difference is two. In row four, the difference isthree and so on. This shows me that there is some sort of relationship between the numerator anddenominator. I can then assume that the equation for the numerator can be used in the equation for thedenominator.Now in order to find the relationship between the position and denominator value I made another chart:Position Number Denominator Value0 151 112 93 94 115 15After a careful blend of logic and luck I finally came up with the second piece of my denominatorequation, which you can see on the next page:
5. 5. n+1 E n (r )=n ( )−r (n−r ) 2The first piece of the equation should be recognizable as the equation for the numerator. The secondpart includes a variable, r, which represents the position within the row of the value being searched for.The variable n did and still does represent the row number.Also, it is important to note that for this equation to work the first position must be treated as positionzero. This allows us to solve for one, which I will explain in more detail when looking at the entireequation.
6. 6. Finding the General EquationOnce we can find the numerator and denominator it is as simple as combining the two equations to findthe entire thing. Of course the numerator equation will be on top and the denominator on bottom, as so: n+1 n( ) 2 E n (r )= n+1 n( )−r ( n−r ) 2The top portion is our numerator equation, and the bottom is simply our denominator equation. Put oneover the other and we have the proper fraction which is present within the triangle.We can find entire rows using this equation, including the initial 1 in each first and last position ineach row. Well find the next two rows, six and seven, using this: 21 21 21 21 21 1 1 16 13 12 13 16 28 28 28 28 28 28 1 1 22 18 16 16 18 22The equation works for both the left and the right side of the triangle.It also works for the ones which begin and end each row, because when r is equal to zero you simplyend up with the numerator over itself, which is of course one.When attempting to find a value past the ending position one ends up with a value lower than 1. This isan easy mistake to catch, therefore we could use this for large rows without much confusion. Simplycalculate until a value under one is found and youve found the end of the row.Unfortunately plugging in each position gets tedious, as even at row seven there are eight differentpositions.Attempting to find negative row numbers simply doesnt work:Row Number Numerator Value-3 3-4 6-5 10-6 15This simply gives us the numerator values for the negative rows turned positive minus one. Rownumber -3 is the same as 2, number -4 is the same as 3, etc.