Solutions 10 MT1100 From Euclid to Mandelbrot 1. The Sierpinski gasket is self-similar, as the n = 3 triangles at stage 1 each become a copy of the Sierpinski gasket, each scaled by a factor of 1/s = 1/2. The fractal dimension is therefore log s log n log(2) log(3) ~ 1.58496. 2. The Menger sponge is obtained by dividing the cube into 33 = 27 smaller cubes of 1/3 the size (in other words, cube scaled by a factor of 1/3) and removing 7 of the 27 cubes. The process is then repeated on the 20 cubes that remain. So the Menger sponge is made up of n = 20 copies of the whole sponge scaled by 1/s = 1/3, and the dimension is therefore log n log s log 20 log 3 ~ 2.72683. 3. Consider the initial triangle ABC below. Drop a perpendicular to AB from C, and let D be the point where AB meets this perpendicular. |AD| = 1/2 and |AC| = 1, and so Pythagoras' Theorem tells us that | AD|2 = 12-(1/2)2 = 3/4 and so |AD| = v3/2. But the area of a triangle of base | AB| and height | AD| is |AB|AD| = v3/4. [A neater way of computing this area is by Heron's formula, if you know this.] 1
U 1 1 D 0 A 1 B At Stage i we add 3 × 4i-1 equilateral triangles to the snowflake, each a copy of the original triangle scaled by a factor of 1/31. Each such copy has area equal to (1/32)2 = 1/92 times the area of the original triangle, so the area of the Koch snowflake is V3 4 (1+ > 00 3 × 41-1 92 i=1 ! . Now 00 00 >(4/9)¿ = (4/9) >(4/9)¿ = (4/9)- 1 1 - (4/9) 5' 4 i=1 i=0 and so the area a is a=13(1+g×1)=25 The length ( of the perimeter of the snowflake is made up of three copies of the Koch curve, which we showed has infinite length. So l is infinite. 4. At the first stage a length 1/5 is removed. At the jth stage, 2j-1 intervals of length (1/5)(2/5)j-1 are removed. So the length L; that remains is 1 Lj=1 -= (1+(4/5)+(4/5)2+ ... +(4/5)j-1) =1- 1 (1 - (4/5)1) 5 (1-(4/5)) = (4/5)j. 2
The middle fifth Cantor set is made up of two scaled copies of itself, each copy scaled by a factor of 2/5. So it has fractal dimension log 2/ log(5/2) ~ 0.756471. Since this is less than 1, we would expect the set to have zero length. Since Lj -> 0 as j -> oo, this is consistent. 3