Math Ib Ia Sl

Jonghyun Choe March 25 2011 Math IB SL Internal Assessment – LASCAP’S Fraction The goal of this task is to consider a set of fractions which are presented in a symmetrical, recurring sequence, and to find a general statement for the pattern. The presented pattern is: Row 1 1 1 Row 2 1 32 1 Row 3 1 64 64 1 Row 4 1 107 106 107 1 Row 5 1 1511 159 159 1511 1 Step 1: This pattern is known as Lascap’s Fractions. En(r) will be used to represent the values involved in the pattern. represents the element number, starting at r=0, and n represents the row number starting at n=1. So for instance, E52=159, the second element on the fifth row. Additionally, N will represent the value of the numerator and D value of the denominator. To begin with, it is clear that in order to obtain a general statement for the pattern, two different statements will be needed to combine to form one final statement. This means that there will be two different statements, one that illustrates the numerators and another the denominators, which will be come together to find the general statement.To start the initial pattern, the pattern is split into two different patterns; one demonstrating the numerators and another denominators. Step 2: This pattern demonstrates the pattern of the numerators. It is clear that all of the numerators in the nth row are equal. For example all numerators in row 3 are 6. 1 1 3 3 3 6 6 6 6 10 10 10 10 10 15 15 15 15 15 15 Row number (n)| 1| 2| 3| 4| 5| Numerator (N)| 1| 3| 6| 10| 15| N(n+1) – Nn| N/A| 2| 3| 4| 5| Table 1: The increasing value of the numerators in relations to the row number.From the table above, we can see that there is a downward pattern, in which the numerator increases proportionally as the row number increases. It can be found that the value of N(n+1) – Nn increases proportionally as the sequence continues. The relationship between the row number and the numerator is graphically plotted and a quadratic fit determ ined, using loggerpro. Figure 1: The equation of the quadratic fit is the relationship between the numerator and the row number. The equation for the fit is: N= 0. 5n2+0. 5n or n2+n2, n>0 Equation 1 In this equation, N refers to the numerator.Therefore, N= 0. 5n2+0. 5n or n2+n2, n>0 is a statement that represents step 2 and also step 1. Step 3: In relation to table 1 and step 2, a pattern can be drawn. The difference between the numerators of two consecutive rows is one more than the difference between the previous numerators of two consecutive rows. This can be expressed in a formula N(n+1) – N(n) = N(n) – N(n-1) + 1. For instance, N(3+1) – N(3) = N(3) – N(2) +1. Using this method, numerator of 6th and 7th row can be determined. To find the 6th row’s value, n should be plugged in as 5 so that N(6) can be found.As for the 7th row’s numerator, n should be plugged in as 6. 6th row numerator is therefore: N(5+1) – N(5) = N(5) – N( 4)+1 N(6) – 15 = 15 – 10+1 N(6) = 15+6 N(6) = 21 7th row numerator is therefore: N(6+1) – N(6) = N(6) – N(5)+1 N(7) – 21 = 21 – 15 +1 N(6) = 42 – 15 + 1 N(6) = 28 Not only by this method, but from the equation found in step 2, figure 1, 6th and 7th row numerator can be found also. 6th row numerator: N(6)=0. 5? 62+0. 5? 6 N(6)=0. 5? 36+3 N(6)=21 th row numerator: N(7)=0. 5? 72+0. 5? 7 N(7)=0. 5? 49+3. 5 N(7)=28 Consequently, these are the values of numerators up to the 7th row. 1 1 3 3 3 6 6 6 6 10 10 10 10 10 15 15 15 15 15 15 21 21 21 21 21 21 21 28 28 28 28 28 28 28 28 Using the method in step 3 and equation 1 in figure 1, it is evident that the numerator in the 6th row is 21. Since both equations have brought same values, it can be concluded that equation 1 is a valid statement that demonstrates the pattern of the numerator.Equation 1 will be used later also, in order to form a general statement of the pattern of whole LACSAP Fract ions. Step 4: When examining the denominators in the LASCAP’S Fractions, their values are the highest in the beginning, decreases, and then increases again. For example, the denominators in row 5 are; 15 11 9 9 11 15. From this pattern, we can easily see that the equation for finding the denominator would be in a parabola form. Element | 0| 1| 2| 3| 4| 5| Denominator | 15| 11| 9| 9| 11| 15|The relationship between the denominator and the element number is graphically plotted and a quadratic fit determined, using loggerpro. Figure 2: This parabola describes the relationship between the denominator and element number. The equation for the fit is : D = r2 – nr+r0 . In this equation, r refers to the element number starting from 0, and r0 being the first denominator value in the row. n refers to the row number starting from 1. To see if this equation work, the 3rd denominator value in the 3rd row was measured. D = 22 – 3 ? 2+6 D = 4 – 6 +6 D = 4With this equat ion, it is evident that the 6th and 7th row denominator values can be found. We already know the first and last denominators from when numerators were found; which are 21 and 28. 6th row second and sixth denominator: D = 12 – 6 ? 1+21 D = 1- 6+21 D = 16 6th row third and fifth denominator: D = 22 – 6 ? 2+21 D = 4- 12+21 D = 13 th row fourth denominator: D = 32 – 6 ? 3+21 D = 9- 18+21 D = 12 7th row second and seventh denominator: D = 12 – 7 ? 1+28 D = 1- 7+28 D = 22 7th row third and sixth denominator: D = 22 – 7 ? +28 D = 4- 14+28 D = 18 7th row fourth and fifth denominator: D = 32 – 7 ? 3+28 D = 9- 21+28 D = 16 Now, since the denominators in the 6th and 7th row are found, the sixth and seventh rows can be drawn and added in the LACSAP’S Fractions. Consequently, these are the fractions up to the 7th row. 1 1 32 1 1 64 64 1 1 107 106 107 1 1 1511 159 159 1511 1 1 2116 2113 2112 2113 2116 1 1 2822 2818 2816 2816 2818 2822 1 Now th at the patterns for the LASCAP’S Fractions are found, all fractions can be expressed in the form En (r) when it is the (r+1)th element in the nth row, starting with r=0. The general statement of the pattern is clearly found when using the equations for the nominator and the denominator.Therefore, the general statement for En r will be En (r) = 0. 5n2+0. 5n r2 – nr+r0 In order to see if the equation works correctly, we can plug in number and figure out if the general statement works out. For example, E7 (3) = 2816 = 0. 5n2+0. 5n r2 – nr+r0 = 0. 5 ? (7)2+0. 5 ? (7) 32 – 7? 3+28 = 2816 . Here, it is clear that the formula is applicable. In order to make sure that the general statement is valid, finding the additional rows of the recurring sequence of fractions by using the general statement above would be useful.Here, I chose to settle on 2 additional rows which are the 8th and 9th rows in the pattern. 8th row numerator: N(8)=0. 5? 82+0. 5? 8 N(8)=0. 5? 64+4 N8=36 9th row numerator: N(9)=0. 5? 92+0. 5? 9 N(9)=0. 5? 81+4. 5 N9=45 8th row second and eighth denominator: D = 12 – 8 ? 1+36 D = 1- 8+36 D = 29 8th row third and seventh denominator: D = 22 – 8 ? 2+36D = 4- 16+36 D = 24 8th row fourth and sixth denominator: D = 32 – 8 ? 3+36 D = 9- 24+36 D = 21 8th row fifth denominator: D = 42 – 8 ? 4+36 D = 16- 24+36 D = 28 9th row second and ninth denominator: D = 12 – 9 ? 1+45 D = 1- 9+45 D = 37 9th row third and eighth denominator: D = 22 – 9 ? +45 D = 4- 18+45 D = 31 9th row fourth and seventh denominator: D = 32 – 9 ? 3+45 D = 9- 27+45 D = 27 9th row fifth and sixth denominator: D = 42 – 9 ? 4+45 D = 16- 36+45 D = 25 Thus, these are the fractions up to the 9th row. 1 1 1 32 1 64 64 1 1 107 106 107 1 1 1511 159 159 1511 1 1 2116 2113 2112 2113 2116 1 1 2822 2818 2816 2816 2818 2822 1 1 3629 3624 3621 3628 3621 3624 3629 1 1 4537 4531 4527 4525 4525 4527 4531 4537 1 This shows th at the general statement for the symmetrical, recurring sequence of fractions is valid and will continue to work.