Calculating Diameters of Interval Unions

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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
 
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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...?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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