Math Ib Ia Sl

Jonghyun Choe March 25 2011 maths IB SL Internal Assessment LASCAPS separate The goal of this task is to consider a set of fractions which argon presented in a symmetrical, hap sequence, and to find a everyday statement for the grade. The presented pose is haggle 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 tint 1 This model is chi green goddessen as Lascaps Fractions. En(r) impart be employ to counterbalance the determine involved in the anatomy. represents the portion itemize, jump at r=0, and n represents the line number first at n=1. So for instance, E52=159, the chip element on the 5th line. Addition completelyy, N impart represent the nurse of the numerator and D value of the denominator. To swallow with, it is clear that in target to prevail a oecumenic statement for the prototype, 2 diametric statements will be call for to combine to make believe one closing statement. This means that there will be tw o different statements, one that illustrates the numerators and another(prenominal) the denominators, which will be come in concert to find the general statement.To start the sign 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 line ar equal. For example all numerators in class 3 argon 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 track number.From the dishearten above, we groundwork see that there is a downward pattern, in which the numerator increases proportionally as the wrangle number increases. It burn be raise that the value of N(n+1) Nn increases proportionally as the sequence come tos. The alliance amidst the course of instruction nu mber and the numerator is diagrammatically plotted and a quadratic tick determined, use loggerpro. find out 1 The comp ar of the quadratic function is the relationship between the numerator and the line number. The comparability for the score 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 yard 2 and also step 1. Step 3 In relation to add-in 1 and step 2, a pattern can be drawn. The difference between the numerators of two consecutive haggles is one more than the difference between the previous numerators of two consecutive course of studys. This can be explicit in a radiation diagram N(n+1) N(n) = N(n) N(n-1) + 1. For instance, N(3+1) N(3) = N(3) N(2) +1. utilize this method, numerator of one-sixth and 7th row can be determined. To find the 6th rows value, n should be blocked in as 5 so that N(6) can be found.As for the seventh rows numerator, n should be blocked in as 6. 6th row numerator is and so 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, calculate 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 atomic number 18 the set 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 utilise the method in step 3 and equation 1 in inning 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 state ment of the pattern of whole LACSAP Fractions. Step 4 When examining the denominators in the LASCAPS 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 slow 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 15The 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 eldest denominator value in the row. n refers to the row number starting from 1. To see if this equation work, the one-third denominator value in the 3rd row was mea sured. 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 stern 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 quaternary 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 LACSAPS 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 that the patterns for the LASCAPS Fractions are found, all f ractions 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 industrial plant 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 one-eighth and 9th rows in the pattern. eighth 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 quaternate 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 that the general statement for the symmetrical, recurring sequence of fractions is valid and will continue to work.