For this problem, our task was to determine
the self-similarity dimension of this shape here.
So first note, it’s the reminder that
this is not exactly the same
as the Sierpinski triangle over here.
So the Sierpinski triangle we remove
a triangle on the middle at every step.
Here in this shape we remove this,
it look like radioactivity symbol to me at every step,
so we remove this three triple triangle.
then from all remain ones
we remove triple triangle and so on.
So end up with this shape.
So what’s its dimension?
Well, we use the familiar formula.
Number of small copies is
magnification factor raise to the dimension D.
So what’s the number of small copies?
So let’s see. 1,2,34,5,6
So I see 6 small copies in here.
And what’s the magnification factor?
Well, that’s 3.
Easiest is to see perhaps along this side,
so here this length
and we have to stretch it 3 times.
Same things to the base,
we would have to stretch it 3 times,
In order for it to be that large,
as the larger shape.
So the magnification factor or the stretch factor is 3.
And then, we need to raise it to the D power.
Alright, so let’s go to work with logs.
Take the log of both sides.
Use the exponent property of logs
to bring the D outside and downstairs,
Then divide both sides by log3
And we’ve done.
log6 over log3
and that’s our dimension.
log6 over log3 there is.
And let’s see, what’s gonna number out of to
an approximate number?
6 log divided by 3 log equals about 1.631
So D approximately 1.631
So that’s the self-similarity dimension of this fractal.