Koch snowflake

Each iteration multiplies the number of sides in the Koch snowflake by four, so the number of sides after ${\displaystyle n}$ iterations is given by:

$${\displaystyle N_{n}=N_{n-1}\cdot 4=3\cdot 4^{n-1}\,.}$$

If the original equilateral triangle has sides of length ${\displaystyle s}$, the length of each side of the snowflake after ${\displaystyle n}$ iterations is:

$${\displaystyle S_{n}={\frac {S_{n-1}}{3}}={\frac {s}{3^{n}}}\,,}$$

an inverse power of three multiple of the original length. The perimeter of the snowflake after ${\displaystyle n}$ iterations is:

$${\displaystyle P_{n}=N_{n}\cdot S_{n}=3\cdot s\cdot {\left({\frac {4}{3}}\right)}^{n}\,.}$$

The Koch curve has an infinite length, because the total length of the curve increases by a factor of ${\displaystyle {\tfrac {4}{3}}}$ with each iteration. Each iteration creates four times as many line segments as in the previous iteration, with the length of each one being ${\displaystyle {\tfrac {1}{3}}}$ the length of the segments in the previous stage. Hence, the length of the curve after ${\displaystyle n}$ iterations will be ${\displaystyle ({\tfrac {4}{3}})^{n}}$ times the original triangle perimeter and is unbounded, as ${\displaystyle n}$ tends to infinity.

As the number of iterations tends to infinity, the limit of the perimeter is:

$${\displaystyle \lim _{n\rightarrow \infty }P_{n}=\lim _{n\rightarrow \infty }3\cdot s\cdot \left({\frac {4}{3}}\right)^{n}=\infty \,,}$$

since ${\displaystyle {\tfrac {4}{3}}>1}$.

An ${\displaystyle {\tfrac {\ln 4}{\ln 3}}}$-dimensional measure exists, but has not been calculated so far. Only upper and lower bounds have been invented. [5]

In each iteration a new triangle is added on each side of the previous iteration, so the number of new triangles added in iteration ${\displaystyle n}$ is:

$${\displaystyle T_{n}=N_{n-1}=3\cdot 4^{n-1}={\frac {3}{4}}\cdot 4^{n}\,}$$

The area of each new triangle added in an iteration is ${\displaystyle {\tfrac {1}{9}}}$ of the area of each triangle added in the previous iteration, so the area of each triangle added in iteration ${\displaystyle n}$ is:

$${\displaystyle a_{n}={\frac {a_{n-1}}{9}}={\frac {a_{0}}{9^{n}}}\,.}$$

where ${\displaystyle a_{0}}$ is the area of the original triangle. The total new area added in iteration ${\displaystyle n}$ is therefore:

$${\displaystyle b_{n}=T_{n}\cdot a_{n}={\frac {3}{4}}\cdot {\left({\frac {4}{9}}\right)}^{n}\cdot a_{0}}$$

The total area of the snowflake after ${\displaystyle n}$ iterations is:

$${\displaystyle A_{n}=a_{0}+\sum _{k=1}^{n}b_{k}=a_{0}\left(1+{\frac {3}{4}}\sum _{k=1}^{n}\left({\frac {4}{9}}\right)^{k}\right)=a_{0}\left(1+{\frac {1}{3}}\sum _{k=0}^{n-1}\left({\frac {4}{9}}\right)^{k}\right)\,.}$$

Collapsing the geometric sum gives:

$${\displaystyle A_{n}=a_{0}\left(1+{\frac {3}{5}}\left(1-\left({\frac {4}{9}}\right)^{n}\right)\right)={\frac {a_{0}}{5}}\left(8-3\left({\frac {4}{9}}\right)^{n}\right)\,.}$$

The limit of the area is:

$${\displaystyle \lim _{n\rightarrow \infty }A_{n}=\lim _{n\rightarrow \infty }{\frac {a_{0}}{5}}\cdot \left(8-3\left({\frac {4}{9}}\right)^{n}\right)={\frac {8}{5}}\cdot a_{0}\,,}$$

since ${\displaystyle {\tfrac {4}{9}}<1}$.

Thus, the area of the Koch snowflake is ${\displaystyle {\tfrac {8}{5}}}$ of the area of the original triangle. Expressed in terms of the side length ${\displaystyle s}$ of the original triangle, this is:[6]

$${\displaystyle {\frac {2s^{2}{\sqrt {3}}}{5}}.}$$

The volume of the solid of revolution of the Koch snowflake about an axis of symmetry of the initiating equilateral triangle of unit side is ${\displaystyle {\frac {11{\sqrt {3}}}{135}}\pi .}$[7]

The Koch snowflake is self-replicating with six smaller copies surrounding one larger copy at the center. Hence, it is an irrep-7 irrep-tile (see Rep-tile for discussion).

The fractal dimension of the Koch curve is ${\displaystyle {\tfrac {\ln 4}{\ln 3}}\approx 1.26186}$. This is greater than that of a line (${\displaystyle =1}$) but less than that of Peano's space-filling curve (${\displaystyle =2}$).

It is impossible to draw a tangent line to any point of the curve.

Graph of the Koch's function

In addition to the curve, the paper by Helge von Koch that has established the Koch curve shows a variation of the curve as an example of a continuous everywhere yet nowhere differentiable function that was possible to represent geometrically at the time. From the base straight line, represented as AB, the graph can be drawn by recursively applying the following on each line segment:

• Divide the line segment (XY) into three parts of equal length, divided by dots C and E.
• Draw a line DM, where M is the middle point of CE, and DM is perpendicular to the initial base of AB, having the length of ${\displaystyle {\frac {CE*{\sqrt {3}}}{2}}}$.
• Draw the lines CD and DE and erase the lines CE and DM.

Each point of AB can be shown to converge to a single height. If ${\displaystyle y=\phi (x)}$ is defined as the distance of that point to the initial base, then ${\displaystyle \phi (x)}$ as a function is continuous everywhere and differentiable nowhere.[3]