Answers to Fractal Math Problems

  
 

Problem 1 -- Area and Length of the Plusses Fractal. This problem has to do with calculating two dimensions of the Plusses fractal, the area of the rhombus as well as the total length of all the lines within it.

Here are the answers to each of the questions
  1. Calculate how the area of the rhombus increases as the fractal grows?

    To calculate the area of the rhombus, you multiple the base times the height. For the first iteration, you can write P(0) = 1 x 1/2. We will now include table for several iterations.
  
 
Iteration
Base
Height
Calculating the area
Area
Growth factor
P(0) 1 1/2 1 x 0.5 0.5 -
P(1)

1+1/2

 1/2+1/4 1.5 x 0.75 1.125 2.25
P(2) 1+1/2+1/4 1/2+1/4+1/8 1.75 x 0.875 1.53125 1.3611
P(3) 1+1/2+1/4+1/8 1/2+1/4+1/8+1/16 1.875 x 0.9375 1.7578125 1.14795918
P(4) 1+1/2+1/4+1/8+1/16 1/2+1/4+1/8+1/16+1/32 1.9375 x 0.96875 1.876953125 1.06777777
P(5) 1+1/2+1/4+1/8+1/16+1/32 1/2+1/4+1/8+1/16+1/32+1/64 1.96875 x 0.984375 1.93798828125 1.03251821
•••••
          
P(¥) 2 1 2 x 1 2 -
 


As you can see the length of the base grows until it reaches the size of 2. The same thing happens to the height that reach the size of 1. So the final area of the rhombus, after many many iteration, ends up being 2.

  1. What happens to the area? Does it keep growing at every iteration or does it stop growing?

    As you can tell from the previous table, the growth of the area slows down on every iteration until it stops at 2.

  2. Now calculate the total length of all the lines inside the fractal at each iteration?

    As you can tell, the size of the the line lengths at the first iteration is 1+1 = 2. We are going to build up a table to show the progression of the total length.

    Iteration Segment length Plusses added Length per plus sign Added length Growth factor Total length
    P(0) 1 - - -   2
    P(1) 1/2 4 3/4 3 2.5 5
    P(2) 1/4 12 3/8 4.5 1.9 9.5
    P(3) 1/8 36 3/16 6.75 1.71 16.25
    P(4) 1/16 108 3/32 10.125 1.62 26.375
    P(5) 1/32 324 3/64 15.1875 1.5758 41.5625
    P(6) 1/64 972 3/128 22.78125 1.54812 64.34375
    P(7) 1/128 2916 3/256 34.171875 1.531083 98.515625

    The Length per plus sign is calculated by dividing the segment length in 2 and then multiplying it by 3. The Added length is calculated by multiplying the Plusses added by the Length per plus sign. Finally, the Total length just adds the Added length to the previous total length.

  3. What happened to the total line length? Does it keep growing at every iteration or does it stop growing?

    The total line length keeps growing on every iteration. The growing factor at around the 52nd iteration it settles at 1.5, and that is why the length keeps growing forever.

  4. Check what happens to the rate of growth of both the area and the total length.

    As described before, the area stops at the size of 2 and the total line length keeps growing and growing.

  5. What do you think about the results? Did you guess that that would happen? Aren't the results interesting?

    I find it very interesting that even though the growth of the area decreases on every iteration, the total length of the lines keeps growing and growing. That is difficult to predict.

Problem 2 -- Total number of +'s on the Plusses Fractal. The next problem that one could work on is to figure out how fast the number of +'s grows. You start with 1, then 5, etc. Finding a formula to predict the total number of +'s for each iteration is not that simple.

Let's start by developing the table.

Iteration Number of line ends Number of +'s to add for this iteration Total number of +'s for this iteration
0 4 1 1
1 12 4 5
2 36 12 17
3 108 36 53
4 324 108 161
5 972 324 485
6 2916 972 1457

The Number of line ends and the Number of +'s to add for this iteration increase by multiplying by 3, so the formula should include 3 to the power of the iteration number. Also, since the Number of line ends starts with 4, it should be multiplied by 4. Here is the table with the formulas

Formulas for total number of plusses
P(0) = 1 P(0) = 1+ 4 (0) P(0) = 1
P(1) = 1+ 4 (1) P(1) = 1+ 4 (1) P(1) = 5
P(2) = 1+ 4 (1+3) P(2) = 1+ 4 (4) P(2) = 17
P(3) = 1+ 4 (1+3+9) P(3) = 1+ 4 (13) P(3) = 53
P(4) = 1+ 4 (1+3+9+27) P(4) = 1+ 4 (40) P(4) = 161
P(5) = 1+ 4 ( 1+3+9+27+81) P(5) = 1+ 4 (121) P(5) = 485
P(6) = 1+ 4 ( 1+3+9+27+81+243) P(6) = 1+ 4 (364) P(6) = 1457

A formula to represent the sequence 0, 1, 4, 13, 40, etc., as shown in the middle column, would be:

(3n - 1) / 2

Where n represents the iteration number. Multiplying this previous formula times 4 and adding 1 results in:

4 x (3n - 1) / 2 + 1 = 2 x (3n - 1) + 1

Which can be simplied to:

Total number of plusses = 2 x 3n - 1

 

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Last Updated: Saturday, 05-Apr-2003 05:33:14 GMT


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