Koch Snowflake

Author

Eric W. Weisstein
January 20, 2017

This notebook downloaded from http://mathworld.wolfram.com/notebooks/Fractals/KochSnowflake.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/KochSnowflake.html.

©2017 Wolfram Research, Inc. except for portions noted otherwise

Additional Authors

Frillflake by Matthew Cook

Graphics

FractalMotif[KochSnowflakeLMotif]

Snowflake

V5

Show[GraphicsArray[KochSnowflake/@Range[0,4]]]

Show[GraphicsArray[KochSnowflakeL/@Range[0,4]]]

V6

GraphicsRow[KochSnowflake/@Range[0,4]]

V11.1

GraphicsRow[Table[Graphics[KochCurve[n]],{n,0,4}]]

Frillflake

Show[GraphicsArray[Partition[Table[KochFrillFlake[3,i,DisplayFunction->Identity],{i,0,3}],2]],DisplayFunction->$DisplayFunction]

Show[GraphicsArray[Partition[Table[KochFrillFlake[4,i,DisplayFunction->Identity],{i,0,3}],2]],DisplayFunction->$DisplayFunction]

Metric Properties

NN[n_]:=3 4^n
L[n_]:=1/3^n
l[n_]:=NN[n]L[n]

Area

FullSimplify[A[n]/.RSolve[{A[n]==A[n-1]+NN[n]L[n]^2Δ/4,A[0]==Δ},A[n],n][[1,1]]
]

FullSimplify[A[n]/.RSolve[{A[n]==A[n-1]+1/3(4/9)^(n-1) Δ,A[0]==Δ},A[n],n][[1,1]]
]

RSolve[{A[n]==A[n-1]+(1/3)((((4/9)))^(n-1))Δ,A[0]==Δ},A[n],n]//FullSimplify

(1/5) (8-(3^(1-2 n)) (4^n)) Δ-(1/5)(8-3((((4/9)))^n))Δ//Simplify

Limit[(1/5)(8-3((((4/9)))^n))Δ,n->∞]