Calculating Diameters of Interval Unions

  • Thread starter Thread starter julia89
  • Start date Start date
  • Tags Tags
    Interval
julia89
Messages
6
Reaction score
0
I know that the diameter for an interval [a,b] is defined as b-a

but what is
1. Diam of (-1,1]U(2,3)
2.Diam of (1,1/2)U(1/4,1/8)U(1/16,1/32)U...

Thanks
 
Physics news on Phys.org
The diameter is defined as the diameter of the smallest interval that contains the set.
 
julia89 said:
I know that the diameter for an interval [a,b] is defined as b-a

but what is
1. Diam of (-1,1]U(2,3)
2.Diam of (1,1/2)U(1/4,1/8)U(1/16,1/32)U...

Thanks
You obviously can't do that unless you know the definition of "diameter of a set or real numbers", not just the diameter of an interval. Fortunately, Dick has provided that.
 
so if i am right

1. Diam of (-1,1]U(2,3)=2+1=3
2.Diam of (1,1/2)U(1/4,1/8)U(1/16,1/32)U...=1/2+1/4+1/16+...

Is this correct?
 
The first one is correct, the second one is not. Did you notice that (1/4, 1/8), (1/16, 1/32), etc. are all SUBSETS of (1, 1/2)? What is (1,1/2)U(1/4,1/8)U(1/16,1/32)U...?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top