Turn on thread page Beta
    • Thread Starter
    Offline

    16
    ReputationRep:
    Hey sexy ladies, how would you go about showing that the numbers made by \displaystyle a_k=k\pi+k^2\pi^2 (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]?

    k\in\mathbb{Z}
    • Wiki Support Team
    Offline

    14
    ReputationRep:
    Wiki Support Team
    What's your definition of 'dense'?
    • Thread Starter
    Offline

    16
    ReputationRep:
    (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.
    • Thread Starter
    Offline

    16
    ReputationRep:
    Anyone?
    Offline

    2
    ReputationRep:
    (Original post by generalebriety)
    What's your definition of 'dense'?
    Totally Tom
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by Hathlan)
    Totally Tom
    That's utterly butterly hilarious.
    Offline

    3
    ReputationRep:
    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???
    • Thread Starter
    Offline

    16
    ReputationRep:
    (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.
    Offline

    1
    ReputationRep:
    (Original post by Hathlan)
    Totally Tom
    :rofl2:
    Offline

    3
    ReputationRep:
    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 ..
    • Thread Starter
    Offline

    16
    ReputationRep:
    (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...?
    Offline

    1
    ReputationRep:
    (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 .
    Offline

    3
    ReputationRep:
    (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??
    Offline

    18
    ReputationRep:
    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.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: January 25, 2010
The home of Results and Clearing

1,284

people online now

1,567,000

students helped last year

University open days

  1. London Metropolitan University
    Undergraduate Open Day Undergraduate
    Sat, 18 Aug '18
  2. Edge Hill University
    All Faculties Undergraduate
    Sat, 18 Aug '18
  3. Bournemouth University
    Clearing Open Day Undergraduate
    Sat, 18 Aug '18
Poll
A-level students - how do you feel about your results?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.