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# showing something is dense watch

1. Hey sexy ladies, how would you go about showing that the numbers made by (mod 2pi) (i.e. taking any value a_k and then shifting it by 2pi until it's in the interval [0,2pi]) are dense in [0,2pi]?

2. What's your definition of 'dense'?
3. (Original post by generalebriety)
What's your definition of 'dense'?
Every point in [0,2pi] is a limit point of the set of points formed by my sequence.
4. Anyone?
5. (Original post by generalebriety)
What's your definition of 'dense'?
Totally Tom
6. (Original post by Hathlan)
Totally Tom
That's utterly butterly hilarious.
7. So whatever point you choose in [0,2pi] it's possible to choose a subset of integers k s.t lim as k-> inf of a(k) = that point???
8. (Original post by ian.slater)
So whatever point you choose in [0,2pi] it's possible to choose a subset of integers k s.t lim as k-> inf of a(k) = that point???
That's what I'm trying to show. Don't know how to do that.
9. (Original post by Hathlan)
Totally Tom
10. I'm rusty/out of my depth here. But you're not getting many takers.

So - I wonder what property pi needs to have to make this conclusion hold? Obviously rationals won't do. That there's only a pi-squared makes me wonder, and I don't see the point of the k*pi at all - surely what's true for one half of the interval will also work for the other?? We need to have lots of ways of choosing subsets, because there's a lot of points to account for. But I guess there's enough ..
11. (Original post by ian.slater)
I'm rusty/out of my depth here. But you're not getting many takers.

So - I wonder what property pi needs to have to make this conclusion hold? Obviously rationals won't do. That there's only a pi-squared makes me wonder, and I don't see the point of the k*pi at all - surely what's true for one half of the interval will also work for the other?? We need to have lots of ways of choosing subsets, because there's a lot of points to account for. But I guess there's enough ..
My reasoning that it is dense is that the pi^2k^2 will never equal the other value. Perhaps if I could show that the distance between the two can be as small as I like then I can make any number...?
12. (Original post by Totally Tom)
My reasoning that it is dense is that the pi^2k^2 will never equal the other value. Perhaps if I could show that the distance between the two can be as small as I like then I can make any number...?
it would be much eaiser if you show the above is bijection of subjection for injection , which can cause rejection .
13. (Original post by Totally Tom)
My reasoning that it is dense is that the pi^2k^2 will never equal the other value. Perhaps if I could show that the distance between the two can be as small as I like then I can make any number...?
Sounds like it ought to be good enough. If for any point p in the interval, and any distance d, you can find a k s.t. |p-a(k)| < d then you can define a sequence of decreasing d's , exhibit the corresponding sequence of k's and then you're OK??
14. I think it's OK (The only possible issue is that I can never remember if you need to be careful about showing that a point actually "hit" by the a(k) sequence also has a convergent series a(n) that tends to the point).

But all that is really doing is rephrasing the original question. The problem is to show that such a k always exists.

I'm not seeing how to do that. The k^2 is a problem. The standard "first step" in these problems is to use the pigeonhole principle to show that for all n, we can find k1, k2 with |a(k1)-a(k2)| < 1/n. Then, if a(k) depends linearly on k, you have |a(k1-k2)| < 1/n, and then for any p in [0,2pi] |a(m(k1-k2)) - p| < 1/n for a suitable choice of m (basically m is the integer part of np).

But that doesn't work with a k^2 term, obviously.

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