# Help understanding Hausdorff dimension of Sierpinski triangle

Discussion in 'General Math' started by Johan Lans, Apr 22, 2007.

1. ### Johan LansGuest

I'm trying to understand this with fractal dimensions, and thought I
got it until I wanted to calculate the dimension of the Sierpinski
triangle.
The formula i use is d=log(N)/log(P/p) , where d is dimesion, N is
the number of units that fit into the larger object, P is the size of
the entire object, and p is the size of the smaller units.
Consider Sierpinski triangle made out of three smaller triangles, the
middle triangle is missing.
The size of P is 1 (one big triangle). The size of a small triangle
is 1/4 of the big one. N is three small triangles that fit into the
bigger one. This gives me d=log(3)/log(4), which is not the correct
answer. Where do i go wrong?

/Johan

Johan Lans, Apr 22, 2007

2. ### Robert IsraelGuest

"Size" here should refer to diameter, not area. The size of a
small triangle is 1/2 the size of the big one.

Robert Israel, Apr 22, 2007

3. ### Johan LansGuest

Robert Israel skrev:
Alright. thanks a lot. Is it the same way when dealing with three
dimensional objects?

/Johan

Johan Lans, Apr 22, 2007
4. ### israelGuest

It's the same in any metric space.

Robert Israel
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

israel, Apr 22, 2007