How to Calculate the Box Counting Dimension in Chaos Theory?

[mathgeek4]

Box counting dimension! please help!

Hi all,

I am working on a problem from Chaos theory, I have to find the box counting dimension of the set {0}U{n^-p} where n is an integer and p>0.
I started this problem by considering p=1. So, the set looks like {0,1,1/2,1/3,...}.
If I take intervals of length 1/2 to cover this set, I need 2 of them.
if I take intervals of length 1/4 to cover this set, I need 3 of them.
This is how far I could go, I am usually able to do the box counting problems easily but this is very confusing.

Please help!

 

[Jamma]

Hmm I think that you can do these by looking at the limit of any decreasing series as epsilon heads to 0.

Notice that for epsilon pretty small, no matter what you do there are going to be some points further from each other than epsilon and hence will need a ball each. So perhaps you can use some sensible series, and perhaps you can then find the number required to cover the "isolated" points and then estimate the number required to cover the rest of the set? Sorry, without working this out I can't give all that detailed advice.

 

[Jamma]

(btw, apparently to wiki, your answer for p=1 is going to be 1/2)

 

[Jamma]

Ok, just gone through it and got the right answer, I'll give you a hint in the right direction, I'm guessing it reasonably easily generalises:

Set epsilon to be 1/(n^2+n). The reason is that (1/n)-(1/n+1)=1/(n^2+n). This tells us that we are going to need at least n balls to cover the first n points (I mean "first" from right to left, i.e. 1, 1/2, 1/3,... up to 1/n).

Now let's look at some estimates for this limit. On the one hand, we could neglect the rest of the set, pretend it's negligible (which, it turns out, it is). So just pretend that it adds nothing, and that the number of 1/(n^2+n)-balls needed is n. If you follow it through, your answer will be 1/2.

So on the other hand, we could give an upper estimate for the rest of the set, and essentially consider it as a portion of the real line, so we'll need to cover [0,n+1]. Clearly a good enough estimate for our limit will be (1/n+1)/(1/n^2+n)=n (i.e. the length of the interval divided by the lengths of the balls). So now our estimate is 2n. This will clearly give the same value in the limit as our lower approximation of n.

Hope this helps.

 

[mathgeek4]

Thanks a lot for your help. I will go through the solution that you gave here tonight, and will get back to you tomorrow. The real challenge is to generalize it for any p.
I appreciate your time.
Thanks.

 

[mathgeek4]

I understand what you did here.
So working on th same lines I tried to generalize it for p=2. If we take the distance between 1/n^2 and 1/(n+1)^2 then it is greater than 1/(n+1)^4. So now if choose this value to be my epsilon, I need atleast n intervals to cover the points 1,1/4,1/9,...
Similarly if I take the difference between 1/n^2 and 1/(n-1)^2 and go from there.

Do you think this would be the correct way to do this?

 

[mathgeek4]

I understand what you did here.
So working on th same lines I tried to generalize it for p=2. If we take the distance between 1/n^2 and 1/(n+1)^2 then it is greater than 1/(n+1)^4. So now if choose this value to be my epsilon, I need atleast n intervals to cover the points 1,1/4,1/9,...
Similarly if I take the difference between 1/n^2 and 1/(n-1)^2 and go from there.

Do you think this would be the correct way to do this?

 

[mathgeek4]

btw the answer for the general case is 1/(1+p).

 

[Jamma]

That looks right to me. Not sure about your "difference between 1/n^2 and 1/(n-1)^2" though, what did you use them for?

And did you check that the rest of the set is still negligible as I did i.e. at each stage, first assume that there are no more balls required to cover the rest of the set and 2nd assume that you have an interval left. I didn't know if this would still work for p<1.

But it looks to me like you've done it right.

 

[mathgeek4]

I think the rest of the would be negligible as n increases. But this way I don't get the correct answer which should be 1/3.
This problem has completely consumed me, it get so complicated when u start using the distance between 1/n^p and 1/(n+1)^p. Do u think I can generalized on similar lines and say that this distance would be greater than 1/(n+1)^2p? If you can from the top of your head give me an estimate for the general case, it would be a great help.
Thanks!

 

[Jamma]

Hmm, you are right, I'm not sure if this method does easily generalise.

I just tried to do the case p>1, and I got a lower bound of 1/2p and an upper bound of 1/2 :S

 

[Jamma]

Right, think I've done it, but had to be pretty liberal with a few limits and things every here and there, but I seem to be getting pretty decisively that the answer should be 1/2p, not 1/(1+p).

 

[mathgeek4]

The answer was given in the book as 1/(1+p). In the general case, did u look at the distance between 1/n^p and 1/(n+1)^p?

 

[mathgeek4]

can you show me how you get 1/2p? Because I was kind of getting to the same thing at one point.

 

[Jamma]

Haha, I've just realized that I must be talking rubbish, it the answer was 1/2p, then for values of p smaller than a half, this dimension would exceed 1, which is of course nonsense. It's weird how the right answer still pops out for the case p=1 though.

Just check that your p can be any real number >0, and not limited to <1 etc. What I had done seemed to have made sense tbh, but I'll have to recheck it, although can't atm bc it's late. I just generalised what I did with the p=1 case and did some manipulations in the limits (as I said, these were quite liberal and hand-wavey at the time, so maybe this is where I went wrong).