Lacsap's fractions - IB portfolio

        k5=0.5×52+0.5×5=12.5+2.5=15

Calculating the numerators of rows 6 and 7 using the general formulae of the above mentioned  patterns:

1.) using the first formula:  kn=2×(n-1)-(n-2)+1

k6= 2×15-10+1=21

k7= 2×21-15+1=28

2.)  using the second formula:  kn=(0.5+0.5n)×(n+1)

k6= (0.5+0.5×6)×(6+1)=3.5×7=21

k7= (0.5+0.5×7)×(7+1)=4×8=28

3.)using the third formula: kn=(n2+n)/2

             k6= (62+6)/2=42/2=21

             k7= (72+7)/2=56/2=28

4.) using the fourth formula:

              k6= 0.5×62+0.5×6=18+3=21

               k7= 0.5×72+0.5×7=24.5+3.5=28

All four  formulae had the same results, thus it can be concluded that they are valid.

GENERAL STATEMENT FOR THE DENOMINATOR

Graph 2: It shows the relation between n (the row number) and mn (the denominator of n). In this case the denominator is the denominator of the outmost fraction of the row (the E0th term, - which is always 1 - is not considered)

This graph is similar to Graph 1, it is also systematically increasing thus the calculations of the denominator’s patterns will be dealt with similarly than those of the numerator are above.

In this part of the portfolio there are 4 major steps which lead to completing the 6th and 7th rows (the steps are numbered) and after those steps comes the complete interpretation for the  general statement of En(r).

1. We already know the numerators of the sixth and seventh rows. Everything so far and the next step are  expanded here:

     1                  1

   1                 3/2                 +2        1

   1                 6/4                 6/4                   +3        1

     1                 10/7                 10/6                 10/7                 +4        1

1               15/11                15/9                  15/9                 15/11          +5         1

      1                     21/16                     21/                              21/                   21/                 21/16   +6          1

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

The numerators are fixed, and they have the same value in each row, since 1 can be written as for example  3/3 or 5/5. Thus, these two patterns can be written as E3(0)=6 and E5(0)=15.

Noticing that the denominator of each row is always one value greater than the difference of the previous two rows’ denominators added to the previous row, the r which in this formula is the (r+1)th’s term denominator is 16 in the 6th row and 22 in the 7th.  

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=mn the denominator of the (r+1)th element can be calculated with r being 0.

2. 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.

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).

3. 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                  

                 4                 4                 

                 7                 6                 7                 

                 11                     9                 9                 11                 

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)

Table 5: It shows the relation of the (r+2)nd and (r+3)rd  denominators of the same row.

Here again the difference between mn(r+2) and mn(r+3) is systematically growing.

Examples:

difference between m6(r+2) and m6(r+3):

m6(r+3)=m6(r+2)-(6-5)→ m6(r+3)=13-(6-5)=12

difference between m7(r+2) and m7(r+3):

m7(r+3)=m7(r+2)-(7-5)→ m6(r+3)=18-2=16

difference between m9(r+2) and m9(r+3):

m9(r+3)=m9(r+2)-(9-5)→ m9(r+3)=31-4=27

Here again it can stated that since the results are the same as those in Table 5 the formula is valid.

However in this case the formula is only valid from the 6thow since that is the first row which consists of three fractions (besides the term 1 in each row).

4. And now completing Lacsap’s fractions with what was calculated in the previous steps and writing down a pattern for the calculation of mn(r+3), which is different from what was investigated on in the previous exercise:

1                 1

1                   +3        3/2                  +3        1

1                 6/4                  +4         6/4                           1

1                 10/7                  +5        10/6                   +5        10/7                         1

1                 15/11                 +6        15/9                   +6        15/9                     +6         15/11                         1    

          1                        21/16            +7   21/13        +7  21/12            +7    21/13           +7    21/16          1

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

Table 6: shows the difference between the denominators of rows  x and (x-2); the difference is the value of the second denominator’s row, row x.

GENERAL STATEMENT FOR En (r)

The probably best formula for the numerator is (n2+n)/2 and such a formula can also be written for the denominator.

When looking for En(r) the example is :

E5(2)=15/9 where 15 can be written as: (n2+n)/2=(52+5)/2=30/2=15

1.)                                           1             1

         2.)                                                    2                  

              3.)                                         4         4                 

         4.)                                                       7           6               7                 

        5.)                                                   11            9              9         11

        6.)                        16               13         12            13             16

               7.)                            22               18           16             16            18            22

This portrayal shows Lacsap’s fraction’s denominators from the first till the seventh row, as does Table 7 to understand the pattern that will lead to the general statement.

Table 7: Comparing the difference between the denominator and the numerator to the row number.

The difference between the numerator and the denominator increases by 1. In this case, the  En(r+1)th term’s denominators are meant, except for the first row, because in that case, it could be both of the 1s.

 It can also be seen that the difference between the numerator and the denominator can be written as :

m(r+2)=k(r+2)+(n-1), where n refers to the row number of the fraction. And the denominator and numerator are meant to be the (r+1)th elements of row n’s fraction.

Following this structure, the difference of the numerator and the denominator in case of the third term (the second fraction) can be written as:

Table 8: It shows the difference between the numerator and the denominator in terms of the row number.

In this case, there is a limitation, as there is no fraction in the first row only the term 1 and thus, the difference cannot be operated on.

In Table 8 it is easy to see that the difference increases by 2 and the formula about this can be written as:

m(r+3)=k(r+3)+(n-2) where n again means the row number and (r+3) means that in this case m and k refer to the third term’s denominator and numerator.

Example:

Considering the expressions:  denominator1=numerator1+(n-1)  and denominator2=numerator2+(n-2),

the general statement for the denominator can be written as:

m=k(n)-r(n-r)  where n means the row number and r means the element number.  Also in this case, the first term means  ’1’ as the first element of the row.

Example:

calculating the denominator of the 3rd term in the 4th row (n=4 and e=3)

m=k(4)-3(4-3)= 10-3=7

calculating the denominators of row 6 and 7:

calculating the denominators of the 6th row:

• if n=6 and r=1:  m=k(6)-1(6-1)= 21-5=16
• if n=6 and r=2:  m=k(6)-2(6-2)= 21-2×4=21-8=13
•  if n=6 and r=3:  m=k(6)-3(6-3)= 21-3×3=9
• if n=6 and r=4:  m=k(6)-4(6-4)= 21-2×4=21-8=13
• if n=6 and r=5:  m=k(6)-5(6-5)= 21-5×1=21-8=16

calculating the denominators of the 7th row:

• if n=7 and r=1:  m=k(7)-1(7-1)= 28-6=22
• if n=7 and r=2:  m=k(7)-2(7-2)= 28-2×5=18
•  if n=7 and r=3:  m=k(7)-3(7-3)= 28-4×3=16
• if n=7 and r=4:  m=k(7)-4(7-4)= 28-3×4=13
• if n=7 and r=5:  m=k(7)-5(7-5)= 28-5×2=18
• if n=7 and r=6:  m=k(7)-6(7-6)= 28-6×1=22

Using this formula, the values of the row are the same as described above:

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

Therefore it can be concluded that both methods are correct.

Knowing the general statement for the numerator and for the denominator a general statement for En(r) can be formulated:

---

En(r)= 0.5n2+0.5n/0.5n2+0.5n-r(n-r)

---

 where n is the row number and r is the element number.

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.