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# Lacsap's fractions - IB portfolio

Extracts from this document...

Introduction

Lacsap’s fractions

MATHEMATICS SL

INTERNAL ASSESSMENT TYPE 1

By: Veronika Kovács

Lacsap’s fractions

INTRODUCTION:

The Lacsap’s fractions are a set of fractions arranged in a symmetrical patter. The task is to consider this set and to try to find a general statement for En(r) with (r+1)th element in the nth  row, starting with r=0. To make the task easier ’ kx ’ will be the sign for the numerator of the xth row and ’mx’ will be the sign for the denominator of the xth row.

1                 1

1                 3/2                  1

1                 6/4                 6/4                 1

1                 10/7                 10/6                 10/7                 1

1                 15/11                     15/9                 15/9                 15/11                  1

PATTERNS OF THE NUMERATOR:

Graph 1 shows that the relation between n and kn is systematically rising:

Graph 1: This graph shows the relation between n (the row number) and kn (the numerator of the nth row.

Since I have found four patterns according to which the numerator can be calculated I have numbered each one to be able to distinguish them better.

1. A  pattern of the numerators can be seen as they are arranged in a row where each term is one value greater than the absolute difference of the previous two numerators added to the previous term.
 kn kn-1 kn-2 2×kn-1-kn-2+1=kn
 6 3 1 2×3-1+1=6
 10 6 3 2×6-3+1=10
 15 10 6 2×10-6+1=15

Table 1: showing the relation between the kn-2th, kn-1th and knth numerators.

As it is shown in table 1 the numerator of the fifth row (k5=15) can be calculated by multiplying the number of the previous row’s numerator  by two (2×10), then subtracting the numerator of the third row (6)

Middle

2×20-15+1=16

Table 3: Shows that using the mn-1th and the mn-2th terms as suggested in  the formula (2× mn-1- mn-2+1=mn), gives the value of mn

so it can concluded that by using the formula:  2× mn-1-mn-2+1=mnthe denominator of the (r+1)th element can be calculated with r being 0.

1. Further adding to the previous  portrayal of Lacsap’s fractions:  m1(r), where r=1, the denominators of rows 6 and 7 can be figured out by the formula:

mn(2)=mn(1)-(n-3)

1                 1

1                 3/2                         1

1                 6/4                 6/4                           1

1                          10/7-1        10/6                           10/7                         1

1                 15/11-2        15/9                15/9-2         15/11                       1

1                                21/16 -3               21/13                          21/                      21/13-3       21/16              1

1                 28/22-4                28/18                    28/                       28/                      28/18-4      28/22         1

Noticing that in the formula mn(r) (with r=1) the denominator of the (r+1)th term and the difference of the denominators of the mn(1)th and the mn(2)nd terms is constantly smaller.  For example in the fifth row the mn(1)th term is 11 and the mn(2)nd term is 9 and in the seventh row the mn(1)th  term is 22 and the mn(2)nd  term is 18.

 row number denominator of the mn(1)th fraction denominator of the mn(2)nd fraction difference of the denominators  within the row
 4 7 6 1
 5 11 9 2
 6 16 13 3
 7 22 18 4

Table 4: Showing the relation of the mn(1)th and mn(2)nd  denominators of the same row.

Seeing that the difference of the denominators is systematically increasing a general formula can be concluded: mn(2)=mn (1)-(n-3)

Example:

calculating the m4(2)th denominator of the fourth row:

m4(2)=m4(1)-(4-3)=7-1=6

calculating the m5(2)th denominator of the fifth row:

m5(2)=m5(1)-(5-3)  m7(2)=11-2=9

calculating the m7(2)th denominator of the seventh row:

m7(2)=m7(1)-(7-3)  m7(2)=22-4=18

The mn(2)nd -s of the examples resulted the same using the formula as they are in Lacsap’s fraction thus the formula is valid.

However this formula has a limitation which is that it can only be applied from the fourth row since that is the first row where there is more than one different type of a fraction (besides the number 1 in each row).

1. The mn(1)th  term of each row can also be figured out by looking at this portrayal where the terms belonging to En(1) and En(2) are the following:

En(1): 1, 2, 4, 7, 11
En(2): 4, 6, 9

1                         1

2

44

7                 67

11                     9                 911

and thus it can be seen that the En(2) sequence can be carried on by the denominators 13 and 18:

1                 1

1                 3/2                         1

1                 6/4                 6/4                           1

1                 10/7                 10/6                 10/7                         1

1                 15/11                     15/9                 15/9                 15/11                   1

1                        21/16               21/13                 21/                            21/13                      21/16             1

1                 28/22                28/18                    28/                     28/                      28/18               28/22           1

And En(3) will be:

En(3)=7,9, 12, 15, 19 etc., because considering the formula (mn(r+2)=mn(r+1)-(n-3)) - with which the mn(r+2)nd element’s denominator can be calculated by knowing  mn(r+1) - a new formula can be made to investigate the mn(r+3)rd  term’s denominator, which is the following:

mn(r+3)=mn(r+2)-(n-5)

 row number mn(r+2) mn(r+3) difference between mn(r+2) and mn(r+3)
 6 13 12 1
 7 18 16 2
 8 24 21 3
 9 31 27 4
 10 39 34 5
 11 48 42 6

Conclusion

And testing the general statement by calculating the 8th and 9th rows using the general statement.

Calculating the 8th row (using GDC)

• if n=8 and r=1:

E8(1)=0.5×82+0.5×8/(0.5×82+0.5×8)-1(8-1)=36/29

• if n=8 and r=2:

E8(2)=0.5×82+0.5×8/(0.5×82+0.5×8)-2(8-2)=36/24

•  if n=8 and r=3:

E8(3)=0.5×82+0.5×8/(0.5×82+0.5×8)-3(8-3)=36/21

• if n=8 and r=4:

E8(4)=0.5×82+0.5×8/(0.5×82+0.5×8)-4(8-4)=36/20

• if n=8 and r=5:

E8(5)=0.5×82+0.5×8/(0.5×82+0.5×8)-5(8-5)=36/21

• if n=8 and r=6:

E8(6)=0.5×82+0.5×8/(0.5×82+0.5×8)-6(8-6)=36/24

• if n=8 and r=7:

E8(7)=0.5×82+0.5×8/(0.5×82+0.5×8)-7(8-7)=36/29

so row number 8 consists of terms, including the ’1’-s since E8(8) and E8(0) also equal 1 :

1              36/29              36/24           36/21           36/20             36/21             36/4                36/29      1

Calculating the 9th row (using GDC)

• if n=9 and r=1:

E9(1)=0.5×92+0.5×9/(0.5×92+0.5×9)-1(9-1)=45/37

• if n=9 and r=2:

E9(2)=0.5×92+0.5×9/(0.5×92+0.5×9)-2(9-2)=45/31

• if n=9 and r=3:

E9(3)=0.5×92+0.5×9/(0.5×92+0.5×9)-3(9-3)=45/27

• if n=9 and r=4:

E9(4)=0.5×92+0.5×9/(0.5×92+0.5×9)-4(9-4)=45/25

• if n=9 and r=5:

E9(5)=0.5×92+0.5×9/(0.5×92+0.5×9)-5(9-5)=45/25

• if n=9 and r=6:

E9(6)=0.5×92+0.5×9/(0.5×92+0.5×9)-6(9-6)=45/27

• if n=9 and r=7:

E9(7)=0.5×92+0.5×9/(0.5×92+0.5×9)-7(9-7)=45/31

• if n=9 and r=8:

E9(8)=0.5×92+0.5×9/(0.5×92+0.5×9)-8(9-8)=45/37

The 9th row of Lacsap’s fraction:

1     45/37       45/31          45/27        45/25        45/25        45/27        45/31         45/37     1

Lacsap’s fraction, the first 9 rows:

1     1

1      3/2        1

1         6/4      6/4          1

1           10/7         10/6         10/7          1

1       15/11             15/9                  15/9           15/11       1

1        21/16         21/13         21/12        21/13      21/16        1

1        28/22      28/18           28/16       28/16          28/18         28/22        1

1         36/29      36/24         36/21          36/20         36/21     36/4       36/29            1

LIMITATIONS

One limitation is that the general statement and all other formulae are only valid if n≥1. For zero or negative n terms it is not valid.

Another limitation is that during the calculations the „1”-s at the beginning and the end of each row did not count when making the calculations. Only in the way that the numerator are divided by the denominator which also equals one, however they could be – and had to be - disregarded at the calculations of the general statement. And thus, the general statement cannot be applied to calculate the elements of the first row.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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