1. Preston Fernandez IB Math SL Internal Assessment March 26, 2011Aim: In this task, individuals will consider a set of numbers that are presented in a symmetricalpattern. This pattern is known as LACSAP’s Fraction and is presented below.Introduction: LACSAP’s Fractions is an array of numbers that are situated symmetrically based upona constant and repeating pattern. Based upon other symmetrical number arrays such as Pascal’striangle, it can be identified that the formation of symmetrical arrays are due to a constant andrepeating patterns between rows. This suggests that the LACSAP’s Fractions array has a constantpattern between the rows and thus an equation of the array can be derived.Finding the 6th and 7th row using patternsNumerator: The differences between the numerators of each row increases by one with every row.This is represented in the figure below. The table clarifies this and anticipates the numerators ofthe 6th and 7th rows. is row number and is the numerator. +2 1 1 +2 2 3 +3 3 6 +3 +4 4 10 +5 5 15 +6 +4 6 21 +7 7 28 +5
2. Preston Fernandez IB Math SL Internal Assessment March 26, 2011Denominator: Like the pattern of the numerators, the difference between each row increases byone, however for denominator it is through elements (the pattern, than with successive rowsbecause denominators change within rows unlike the numerators. This is shown below. Element Number 0 1 2 3 4 5 6 7 5 1 11 9 9 11 1 P +5 +4 +3 +2 6 1 16 13 12 13 16 1 +2 P +6 +5 +4 +3 +2 7 1 22 18 16 16 18 22 1 The table above attempts to continue the +3 +2 pattern left off from the figure to the left. is row number, and P shows how the +4 pattern would affect the denominator +3 +2 between elements of different rows.Continuing the pattern of these the 6th and 7th rows come out to beDeriving the Equation for the Numerator thLet , where is row number and is the numerator of the rowBy allowing the relationship between row number and the numerator of that row can beidentified. The table below represents this relationship. 1 1 2 3 3 6 4 10 5 15Patterns: Relationship between Row Number and NumeratorThere is a geometric relationship between the rows and the numerators. As row number increases,the number multiplied for it to become equal to the numerator increases by 0.5 each individual row.This is shown in the table below. Let be equivalent to the amount that multiplies to equal . 1 .0 1 2 3 3 6 4 10 5 15
3. Preston Fernandez IB Math SL Internal Assessment March 26, 2011Another pattern was derived from the pattern previously investigated. It found that when wasdoubled, and then was multiplied by , the value of was twice as great. An equation thatrepresents this more clearly is . More importantly, it was identified that 2 is alsoequivalent to . An equation that represents this is . This is expressed in the tablebelow. 1 .0 1 1 2 2 2 3 2 3 6 3 6 3 4 12 4 10 4 5 20 5 15 5 6 30 +1So because and , can be substituted in for . This gives the equationof which can be further simplified into . is the equation used to find the numerator.Validating the Equation for the Numerator Using Patterns and GraphsThe equation used to find the numerator can also be found by plotting the data points and findingthe line of best fit.Patterns: Relationship between Numerators of Successive RowsThe differences between numerators does not remain constant between each row; rather, thedifference of numerators between each row is increased by one. The difference between row twoand row one is two, the difference between row three and row two is three, the difference betweenrow four and row three is four and so on. This is more explicitly shown in the table below. 1 1 +2 2 3 +3 3 6 +4 4 10 +5 5 15To derive the equation from the graph that this pattern plots, more data points will be needed, sobased on this pattern the sequence will be continued and then graphed when there is a reasonableamount of data points. This ensures that the equation derived earlier applies to all of the array andthe pattern, and not just the given section.
4. Preston Fernandez IB Math SL Internal Assessment March 26, 2011Row Numerator 1 1 Row Vs. Numerator 2 3 3 6 350 4 10 300 y = 0.5x2 + 0.5x - 2E-13 R² = 1 5 15 250 Numerator 6 21 200 7 28 150 Numerator 8 36 100 Poly. (Numerator) 9 45 50 10 55 0 11 66 0 10 20 30 12 78 Row Number 13 91 14 105 15 120 The graph above represents how row number affects the numerator. As 16 136 row increases, so does it numerator, however it is not in a linear fashion. 17 153 The difference from one row to another increases the row count gets 18 171 larger. 19 190 20 210 The equation calculated from the graph is identical to the equation 21 231 derived however this equation shows the y-intercept. 22 253 23 276 The equation 0.5x2 0.5x is the same equation as however has 24 300 been simplified out to 0.5x2 0.5x. 25 325 This validates and supports the derived equation.Validating the Equation for the Numerator by Plugging InThe equation can be validated by plugging in given numbers, which pertain to the first 5 rows of theLACSAP Fractions array. This is done by substituting numbers from the given information into theequation to see if the end product and the given numerators are equal. Plugging in for validation isgiven below, where is row number and is the numerator of that row. Given Information 1 1 2 3 3 6 4 10 5 15When the given information is substituted into the equation, it is seen that the end solution isequivalent to that of the numerator which suggests that this equation can be used to find thenumerator of each row.
5. Preston Fernandez IB Math SL Internal Assessment March 26, 2011Deriving the Equation for the DenominatorLet’s compare the numerator and denominator of the second element in each row.In the table below is the row number, is the value of the numerator, is the element numberand is the denominator. Note* that the first term in each row is . 1 1 1 1 2 3 1 2 3 6 1 4 4 10 1 7 5 15 1 11From this table a pattern was seen that . This is verified below. 1 1 1 1 3 2 1 2 6 3 1 4 10 4 1 7 15 5 1 11Although the algebra is true, this equation only is fit for , the first element. This is verifiedbelow when the equation applies to the second element or the third term of the row. 3 2 2 3 6 3 2 4 10 4 2 6 15 5 2 9Then it was questioned why the formula had worked for the first element (2nd term) and not thesecond element (3rd term), so the values plugged into the formula were reevaluated and looked atmore closely to determine why the formula worked for the first element and not the others. In thisreassessment, was looked at as a single unit rather than the difference of two values. This wasdone to assure nothing had been overlooked as well as identify any new patterns which may havebeen applicable.
6. Preston Fernandez IB Math SL Internal Assessment March 26, 2011In the table below let , as mentioned previously this is done to assure nothing had beenoverlooked as well as identify any new patterns which may be applicable. 3 2 2 3 6 3 2 4 10 4 2 6 15 5 2 9From the table above it was seen that was too small of a value and when subtracted from andhence the difference was higher than the given value, . A pattern was also seen that when wasdoubled, and then subtracted from , the difference had equaled the value of . And from this theequation was derived. Now this equation only works for the seconds element and nownot the first, the third or any other. It was then seen that the coefficient must equal the elementnumber and hence the equation for all the denominators in every element and row was derived as or more formally as .For this equation to work, the numerator must be known, which as previously found wasSo the equation for the denominator isValidating the Equation for the Numerator by Plugging InEquation will be tested against all elements of row five. is the element number, is the row number, and is the given value of the denominator 1 11 2 9 3 9 4 11When the given information is substituted into the equation, it is seen that the end solution isequivalent to that of the given denominator which suggests that this equation can be used to findthe denominator of each element of any row.
7. Preston Fernandez IB Math SL Internal Assessment March 26, 2011Finding additional rows using the derived equations is the equation for both numerator and denominator of a term. n r 0 1 2 3 4 5 6 7 8 9 10 1 1 1 2 1 1 3 1 1 4 1 1 5 1 1 6 1 1 7 1 1 8 1 1 9 1 1 10 1 1The table above shows the first 10 terms in LACSAP’s Fractions Array. The equationwas used to get all the values above, and also validates and verifies the formula derived.
8. Preston Fernandez IB Math SL Internal Assessment March 26, 2011Scope and Limitations of the General Statement must be greater than 0 at all times. If were not greater than 0, the solution or denominator ofan element maybe undefined. As well there cannot be a “0th” row in the pattern as it cannot exist.If were a negative, and substituted into the equation, the answer would become positive, where itwould actually need to be negative. For these two reasons, the equation restricts from being lessthan 0, and hence for the equation to work must be greater than 0. As well rows cannot benegative because there cannot be “negative rows”. Must be greater than 0 has there cannot be 0 elements in a pattern. For a pattern to be a pattern,there must be a range, if there are 0 elements, there are 0 numbers. must be a real number, and not a fraction. There cannot be “half” elements and hence must bea real number. must also be positive as there cannot be negative elements.