一條中六既數學問題...

圖片參考：http://upload.wikimedia.org/wikipedia/commons/thumb/d/d9/KochFlake.svg/280px-KochFlake.svg.png


Taking s as the side length, the original triangle area is
圖片參考：http://upload.wikimedia.org/math/e/6/c/e6c1df0c94756d5cef1bff73312f3289.png
triangles. Combining these two formulae gives the iteration formula:


圖片參考：http://upload.wikimedia.org/math/8/a/a/8aa6d7febcc61f86c64450536a7ef5c9.png

where A0 is area of the original triangle. Substituting in


圖片參考：http://upload.wikimedia.org/math/9/5/6/9569f22af8487bd8fa8f7fb7257fd530.png

and expanding yields:


圖片參考：http://upload.wikimedia.org/math/3/e/b/3eb15f1fe36a16cbe1cc01cd53f576c7.png

In the limit, as n goes to infinity, the limit of the sum of the powers of 4/9 is 4/5, so


圖片參考：http://upload.wikimedia.org/math/e/8/c/e8cc19c0eef23aa55711feb6732398ce.png

So the area of a Koch snowflake is 8/5 of the area of the original triangle, or
圖片參考：http://upload.wikimedia.org/math/5/1/6/51662cacaf689eb28029a4a2a5582acf.png
.[2] Therefore the infinite perimeter of the Koch triangle encloses a finite area.


中文解釋可參考 :

http://www.math.ied.edu.hk/ITProj2003/Module_4/Aesthetic_Perspective.htm

2010-06-03 13:03:40 補充：
The Koch curve has an infinite length because each time the steps above are performed on each line segment of the figure there are four times as many line segments, the length of each being one-third the length of the segments in the previous stage.

2010-06-03 13:03:45 補充：
Hence the total length increases by one third and thus the length at step n will be (4/3)n of the original triangle perimeter.

2010-06-03 13:14:43 補充：
Here S = 1 cm ,

the area = (2√3)/5 cm^2

Firstly, let N(n) be the no of sides of the shape after it changes n times and as we know that N(0)=3

Now, we note that every side will produce 4 sides, therefore

N(1)=4(3) =12
N(2)=4．N(1)=48

therefore N(n)=3．4^(n)

Then Let L(n), S(n) be the length and the perimeter of the shape after n-th change respectively

we can observe that L(n)=1．(1/3)^(n)
therefore
S(n)=L(n)．N(n) = (1/3)^(n)．(3)．4^n
=3．(4/3)^n

as n tends to infinity, (4/3)^n also tends to infinity, then the perimeter of the shape also tends to infinity.


Let A(n) be area of shape after n-th change

A(0) = (1/2)(1)(sin 60) =(1/2)(√3/2)=√3/4
we can see that every side will produce one more triangle with area (1/9)^n * A(0)
hence we have
A(1)=A(0) + N(1)*(1/9)A(0)=A(0)+3*4(1/9)A(0)
A(2)= A(0)+3*4(1/9)A(0)+3*4^2*(1/9)^2A(0)

A(n)=A(0)+ 3*A(0)*[ 4/9+(4/9)^2+(4/9)^3+....+(4/9)^n)

we can see that 4/9+(4/9)^2+...+(4/9)^n is the sum of the geometric series with ratio of 4/9

therefore we have
A(n)=A(0)+3*A(0)*[(1-(4/9)^n)/(1-4/9)]

as n tends to infinity
we have
A(∞)=A(0)+3*A(0)[9/5]
=√3/4(1+27/5]
=√3(8/5)


希望幫到你！


2010-06-04 18:23:21 補充：
對不起！最後運算錯誤
應該是

A(1)=A(0) + N(０)*(1/9)A(0)=A(0)+3*(1/9)A(0)
A(2)= A(0)+3*（１）(1/9)A(0)+3*4*(1/9)^2A(0)

2010-06-04 18:23:37 補充：
A(n)=A(0)+ 3*A(0)*[ 1/9+(4)(1/9)^2+(4)^2(1/9)^3+....+(4)^n-1*(1/9)^n)

=A(0)+(3/4)*(4/9)A(0)[1+4/9+(4/9)^2+......+(4/9)^n]
we can see that 1+4/9+(4/9)^2+...+(4/9)^n is the sum of the geometric series with ratio of 4/9

2010-06-04 18:23:43 補充：
therefore we have
A(n)=A(0)+(1/3)*A(0)*[(1-(4/9)^n)/(1-4/9)]

as n tends to infinity
we have
A(∞)=A(0)+(1/3)*A(0)[9/5]

=√3(2/5)