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This process is repeated indefinitely. Figure 2.1 the first two steps of this con-
                             struction. It is interesting to observe that the Koch curve is an example of a
                             curve where there is no way to fit a tangent to any of its points. In a sense, it
                             is an example of a curve that is made out of corners everywhere.
                              The detailed study of these objects is covered in courses in fractal geometry,
                             chaos, dynamic systems, etc. We limit ourselves here to the simple problems
                             of determining the number of segments, the length of each segment, the
                             length of the curve, and the area bounded by the curve and the horizontal
                                              th
                             axis, following the k step:
                                1. After the first step, we are left with a curve made up of four line
                                   segments of equal length; after the second step, we have (4 × 4)
                                   segments; and the number of segments after k steps, is

                                                           n(k) = 4 k                      (2.11)

                                2. If the initiator had length L, the length of the segment after the first
                                                  2
                                   step is L/3, L/(3) , after the second step and after k steps:
                                                         s(k) = L/(3) k                    (2.12)

                                3. Combining the results of Eqs. (2.11) and (2.12), we deduce that the
                                   length of the curve after k steps:

                                                                    k
                                                        Pk() =  L ×    4                 (2.13)
                                                                   3

                                4. The number of vertices in this curve, denoted by u(k), is equal to
                                   the number of segments plus one:

                                                                k
                                                         u(k) = 4  + 1                     (2.14)
                                5. The area enclosed by the Koch curve and the horizontal line can be
                                   deduced from solving a difference equation: the area enclosed after
                                                                                 th
                                       th
                                   the k  step is equal to the area enclosed in the (k – 1)  step plus the
                                   number of the added triangles multiplied by their individual area:
                                                                     uk() −  uk( − 1 )
                                          Number of new triangles =                        (2.15)
                                                                         3      


                                                                                   
                                                                       3  2      3 1  2k  2
                                   Area of the new equilateral triangle =   sk() =      L  (2.16)
                                                                                    3
                                                                       4        4  


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