Maths SL TYPE I Lacsap’s Fractions Portfolio Lacsap’s fraction Introduction: Lacsap’s fraction is a symmetrical triangle that has the following pattern in the first five rows The shape is similar to Pascal triangle. It has the same quantity of symmetrical triangle as Pascal triangle. And Lacsap is the inverse alphabet order of Pascal. These make me think about Pascal triangle and I made an assumption that elements in Lacsap triangle may have the same relationship as in Pascal triangle.However, the elements in Lacsap’s fraction triangle may or may not have the same relationship as Pascal triangle so I ignore my hypothesis about Pascal triangle and decide to find the relationship by not referring to Pascal triangle. I believe that every element in Lacsap’s triangle must be in a sequence and the task for this portfolio is to find the relationship between each element. This portfolio will be divided into five parts; finding numerator relationship, finding denominator relationship, test the accuracy of the statement, finding additional rows and limitations of the statement. The notations in this portfolio are X = the element place * N = row number * En(x) = The xthelement on the nthrow Numerator Relationship: First thing I notice about the Lacsap is that the numerators are the same in the same row. The numerators are the same in each row My first statement about Lacsap’s fraction is the numerators are the same in each row. Considering the second diagonal row may give me some clue about how to find the numerator. If I can find the numerator for the second diagonal, the numerator for that row is revealed because the numerators are the same in the same horizontal row. Second diagonal row Row number| Numerator| 1st difference| 2nd difference| | 1| -| -| 2| 3| 2| -| 3| 6| 3| 1| 4| 10| 4| 1| 5| 15| 5| 1| I can see that the numerator’s 1st difference is 2, 3, 4 and 5 for the first five rows and the 2nd difference is 1. The numerators are triangular number. Please see the next page for the explanation of triangular number. Triangular numbers are a sequence of number that follows the following pattern: Term| Triangle| Number| 1| | 1| 2| | 3| 3| | 6| 4| | 10| And so on…. The table below shows how the numerators in each row are derived. Row number| Numerator| Numerator| 1| 1| 1| 2| 1+2| 3| 3| 1+2+3| 6| 4| 1+2+3+4| 10| 5| 1+2+3+4+5| 15|Plot the numerators in the graph to find the relationship between row number and the numerator. n= row number| f(n)= Numerator| 1| 1| 2| 3| 3| 6| 4| 10| 5| 15| I used Casio FX9860 Calculator to plot the graph and find the equation. Given that f(n) = row number and n= numerator. Plot Data input Graph Equation The equation of the graph isfn=0. 5n2+0. 5n Therefore, I can make a statement about finding the numerator that: numerator=0. 5n2+0. 5n numerator=nn+12 Where: n=row number Conclusion for finding the numerator: * All the numerators are the same in the same row. * numerator=nn+12 when n = row number Denominator Relationship:Turning the first and last element of each row (they are 1 for every row) to fraction may give me some clue of how to find the denominator. The numerator and denominator of the fraction will be the same to make the fraction equal to 1. From the generalisation I have made before “the numerators are the same in each row” the numerator of the first and last element will equal to the numerator of the other elements in that row. This is a new Lacsap triangle. The denominators in each row seem to have a quadratic relationship. I think this is because first and last denominators of each row are the highest and the middle denominator is the lowest.The denominators are symmetrical which means everything on the left half is equal to the right half. I used Logger Pro to find the relationship of the denominators in each row. I skipped finding row 1 because there are insufficient values in row 1. There are only two denominators; 1 and 1. These denominators are not enough to draw an accurate graph. Row 2 x= element place| f(x)= denominator| 1| 3| 2| 2| 3| 3| fx=x2-4x+6 Row 3 x= element place| f(x)= denominator| 1| 6| 2| 4| 3| 4| 4| 6| fx=x2-5+10 Row 4 x= element place| f(x)= denominator| 1| 10| 2| 7| 3| 6| 4| 7| 5| 10| fx=x2-6x+15 Row 5 x= element place| f(x)= denominator| | 15| 2| 11| 3| 9| 4| 9| 5| 11| 6| 15| fx=x2-7x+21 I’ve found that the relationship between the element place and the denominator is given on the following table: Row| Equation| 1| Skipped| 2| fx=x2-4x+6 (from page 6)| 3| fx=x2-5x+10 (from page 7)| 4| fx=x2-6x+15 (from page 8)| 5| fx=x2-7x+21 (from page 9)| Where f(x) = the denominator and x = the element place Using the above table helped me to see how the functions on the right hand column relate to the others. The relationship that I have found will be shown on the following Given that the denominators in each row have the following relationship: fx=ax2+bx+c A is always equal to + 1 * B is(-|row number|)– 2 * Let row number = n * B = (-|n|)-2 * Cs are triangular numbers that start with 6 on the second row * The relationship is shown on the next page Equation 1 At this point I have found that fx=x2-(n+2)x+c When f(x) = the denominator, x = element place and n = row number Note this equation as equation 1. It’ll be used later once I’ve found C. Finding C: x = Row number| f(x) = C| 2| 6| 3| 10| 4| 15| 5| 21| Using Logger Pro to find the relationship C can be found by using the following equation Equation 2 fn=0. 5n2+1. 5n+1 When n = row number and f (n) = CNote this equation as equation 2. It’ll be used to combine with the equation 1. Equation 1 on page 10 Therefore, the denominators follow the relationship of: fx=x2-(n+2)x+c Equation 2 on page 11 C=f(n)=0. 5n2+1. 5n+1 Substitue C from equation 2 to equation 1 fx=x2-n+2x+(0. 5n2+1. 5n+1) Denominator=x2-n+2x+(0. 5n2+1. 5n+1) When x = element place and n = row number Conclusion for finding the denominator: * The denominators in the same row have a quadratic relationship for every row. They can be plotted as a parabolic shape curve. * The denominators are symmetrical * The equation to find the denominator is Denominator=x2-n+2x+(0. n2+1. 5n+1) When x = element place and n = row number Let Enx be the xth element in the nth row. Enx=Numerator equation on page 4Denominator equation on page 13 Enx=nn+12×2-n+2x+(0. 5n2+1. 5n+1) Enx=nn+12(x2-n+2x+(0. 5n2+1. 5n+1)) Enx=nn+12(x2-nx-2x+0. 5n2+1. 5n+1) Enx=nn+12×2-2nx-4x+n2+3n+2 Test the accuracy of this statement Enx=nn+12×2-2nx-4x+n2+3n+2 Test 1: Row n = 5Element x = 6 Enx=nn+12×2-2nx-4x+n2+3n+2 E56=55+12(6)2-2(5)(6)-4(6)+52+3(5)+2 E56=3072-60-24+25+15+2 Correct E56=3030=1 Test 2: Row n = 3Element x = 3 Enx=nn+12×2-2nx-4x+n2+3n+2 E33=33+12(3)2-2(3)(3)-4(3)+32+3(3)+2 E33=12(18)-(18)-12+9+9+2 CorrectE33=128=64 Test 3: Row n = 1Element x = 2 Enx=nn+12×2-2nx-4x+n2+3n+2 E12=11+12(2)2-2(1)(2)-4(2)+12+3(1)+2 E12=28-4-8+1+3+2 Correct E12=22=1 Test 4: Row n = 5Element x = 3 En(x)=nn+12×2-2nx-4x+n2+3n+2 E53=55+12(3)2-2(5)(3)-4(3)+52+3(5)+2 E53=5618-30-12+25+15+2 E53=3018=159 Correct The formula En(x)=nn+12×2-2nx-4x+n2+3n+2 is accurate to find the elements in Lacsap’s fractions. Finding the next rows: The elements in additional rows will be separated into two parts; numerators and denominators. Row 6 Numerator: Given that n = row number Numerators=nn+12 Numerators in row 6=66+12 Numerators in row 6=422 Numerators in row 6=21 2121212121 1 Row 6 Denominator: Given that f(x) = denominator, x = element place and n = row number. Denominator=x2-n+2x+(0. 5n2+1. 5n+1) Denominator in row 6=x2-8x+28 fx=x2-8x+28 N = row number| X = element place| F(x) = denominator| 6| 1| 21| 6| 2| 16| 6| 3| 13| 6| 4| 12| 6| 5| 13| 6| 6| 16| 6| 7| 21| Row 6 Therefore, elements in sixth row are: 1 21162113211221132116 1 Row 7 numerators: Given that n = row number Numerators=nn+12 Numerators in row 7=77+12 Numerators in row 7=562 Numerators in row 7=28 1 282828282828 1 Row 7 denominators: Given that f(x) = denominator, x = element place and n = row number.Denominator=x2-n+2x+(0. 5n2+1. 5n+1) Denominator in row 7=x2-9x+36 I used Autograph to draw this function and see what the denominators are. N = row number| X = element place| F(x) = denominator| 7| 1| 28| 7| 2| 22| 7| 3| 18| 7| 4| 16| 7| 5| 16| 7| 6| 18| 7| 7| 22| 7| 8| 28| Row 7 Therefore, the elements in this row are 1 282228182816281628182822 1 Row 8 Numerator: Given that n = row number Numerators=nn+12 Numerators in row 8=88+12 Numerators in row 8=36 1 36363636363636 1 Row 8 Denominator: Given that f(x) = denominator, x = element place and n = row number. Denominator=x2-n+2x+(0. 5n2+1. 5n+1)Denominator in row 8=x2-10x+45 N = row number| X = element place| F(x) = denominator| 8| 1| 36| 8| 2| 29| 8| 3| 24| 8| 4| 21| 8| 5| 20| 8| 6| 21| 8| 7| 24| 8| 8| 29| 8| 9| 36| Row 8 Therefore, the elements in this row are 363636293624362136203621362436293636 Row 9 Numerator: Given that n = row number Numerators=nn+12 Numerators in row 9=99+12 Numerators in row 9=45 1 4545454545454545 1 Row 9 Denominator: Given that f(x) = denominator, x = element place and n = row number. Denominator=x2-n+2x+(0. 5n2+1. 5n+1) Denominator in row 9=x2-11x+55 N = row number| X = element place| F(x) = denominator| 9| 1| 45| 9| 2| 37| | 3| 31| 9| 4| 27| 9| 5| 25| 9| 6| 25| 9| 7| 27| 9| 8| 31| 9| 9| 37| 9| 10| 45| Row 9 Therefore, the elements in this row are 1 45374531452745254525452745314537 1 Row 10 Numerator: Given that n = row number Numerators=nn+12 Numerators in row 10=1010+12 Numerators in row 10=55 1 555555555555555555 1 Row 10 Denominator: Given that f(x) = denominator, x = element place and n = row number. Denominator=x2-n+2x+(0. 5n2+1. 5n+1) Denominator in row 10=x2-12x+66 N = row number| X = element place| F(x) = denominator| 10| 1| 55| 10| 2| 46| 10| 3| 39| 10| 4| 34| 10| 5| 31| 10| 6| 30| 10| 7| 31| 10| 8| 34| 10| 9| 39| 10| 10| 46| 0| 11| 55| Row 10 Therefore, the elements in this row are 1 554655395534553155305531553455395546 1 Row 1 This is the first ten rows of Lacsap’s fractions: 1 1 Row 2 1 32 1 Row 3 1 6464 1 Row 4 1 107106107 1 Row 5 1 15111591591511 1 Row 6 1 21162113211221132116 1 Row 7 1 28 2228182816281628182822 1 Row 8 1 3629362436213620362136243629 1 Row 9 1 45374531452745254525452745314537 1 Row 10 1 554655395534553155305531553455395546 1 The limitations of Enx Enx=nn+12×2-n+2x+(0. 5n2+1. 5n+1) I’ve noticed that the number of elements in a row is equal to the row number + 1, X ? n+1.I wonder what would happen if x > n+1. I chose x=6 n=4 to see the result when x > n+1. Given that x = 6 and n = 4 Enx=44+12(6)2-4+2(6)+(0. 542+1. 5(4)+1) E46=44+12(6)2-4+2(6)+(0. 542+1. 5(4)+1) E46=45236-6(6)+(0. 516+6+1) E46=1036-36+8+6+1 E46=1015 When x > n+1, the element is less than one, while every element in Lacsap’s fractions is greater or equal to one. Other limitations of Enx is that 0? x2-n+2x+0. 5n2+1. 5n+1 and 0? 2×2-2nx-4x+n2+3n+2 since the denominator must not equal to zero Conclusion: 1. All the numerators are the same in the same row 2. The first and last elements are always 1 3. The fractions are symmetrical . Number of elements in a row is equal to row number + 1. Number of elements = row number +1 a. Given that x = element place and n = row number, x? n+1 5. Given that x = element place and n = row number b. Numerator=nn+12 (from page 4) c. Denominator=x2-n+2x+0. 5n2+1. 5n+1 (from page 12) d. Enx=nn+12×2-n+2x+(0. 5n2+1. 5n+1)OR e. Enx=nn+12×2-2nx-4x+n2+3n+2(from page 13) The conditions of Enx formula are: * x2-n+2x+0. 5n2+1. 5n+1? 0 because the denominator cannot equal to 0 * 2×2-2nx-4x+n2+3n+2? 0 because the denominator cannot equal to 0 * X ? n+1. If X ? n+1, the element will be outside Lacsap’s fractions.