Page 2 of 3 FirstFirst 123 LastLast
Results 16 to 30 of 36

Math Help - Density in the unit circle

  1. #16
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by Hartlw View Post
    I sense FernandoRevilla is correct though I can't follow it (I am not qualified to judge). Defunkt states, in effect, a countably infinite set of points in a real interval doesn't include all the points. That is not the question at hand, though I admit I was a little sloppy in some previous phrasing about "all points."

    From an Engineering perspective:

    e^ik is a point on the unit circle at an angle of k radians.

    If I proceed around the unit circle in steps of k radians do I ever hit the same point again, ie, after m steps do I move a multiple of 2pi? Or, does k exist such that m=2pik? Is pi rational? No.
    So far, correct.

    So I have a countably infinite number of points on the circle. Are they dense?

    Somehow this has to get to the equivalent question, are the rational points on [0,1] dense? Yes, because every real point of the interval is a limit point of the rational numbers on the interval.
    I don't see how this question reduces to the density of the rationals in [0,1] - surely not every dense set in [0,1] is \mathbb{Q} \cap [0,1], right? I mean, for example, cos(1) is not rational..

    An infinite number of points on the circle must indeed have a limit point, but is every point hit as per above a limit point, in terms I could understand? In other words, given any point on the circle, are there points constructed as per above on the circle within an arc length epsilon?

    EDIT: So I have the question, given any point on the unit circle, how many times do I have to go around to get to within a distance epsilon? Getting close.
    Fernando's proof essentially seals the deal. What do you not understand about it?
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98
    Sorry you missed my edit. Let me summarize:

    e^ik is a point on the unit circle at an angle of k radians.

    If I proceed around the unit circle in steps of k radians do I ever hit the same point again, ie, after m steps do I move a multiple of 2pi? Or, does k exist such that m=2pik? Is pi rational? No.

    So I have a countably infinite number of points on the circle. Are they dense?

    An infinite number of points on the circle must indeed have a limit point, but is every point hit as per above a limit point, in terms I could understand? In other words, given any point on the circle, are there points constructed as per above on the circle within an arc length epsilon?

    EDIT: So I have the question, given any point on the unit circle, how many times do I have to go around to get to within a distance epsilon? Getting close.

    OK: For any angle a and any desired degree of accuracy (couched in terms of epsilon),
    m/n<=a<=(m+1)/n
    m<=na<=(m+1)
    k=m

    I don't see where Fernando proves you can get to within a distance epsilon of any point other than "..there must be integers m and n ..."
    But pick whichever proof you like.
    Follow Math Help Forum on Facebook and Google+

  3. #18
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    I don't really understand your proof. From what I gather, you say that there exist integers m,n \in \mathbb{N} such that \frac{m}{n} \le a \le \frac{m+1}{n}, but what is k? and what does choosing k=m prove?

    Fernando showed that for any \epsilon > 0, you can 'cover' the circle with arcs of length smaller than \epsilon, each of them having a point of the form e^{ik} inside. Do you see why this proves the result?
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98
    From Rosenlicht, Intro to Analysis, pg26, LUB4

    LUB 4. For any x in R and positive integer N, there is an integer n st n/N<=x<(n+1)/N

    OK: For any angle a and any desired degree of accuracy (couched in terms of epsilon),
    m and n exist st (see Rosenlicht above):
    m/n<=a<(m+1)/n
    m<=na<(m+1)
    k=m

    If e^ik = e^im, then k/n<=a<(k+1)/n
    ie, e^ik differs from e^ia by at most 1/n, n arbitrary..

    e^ia is the point on the unit circle at angle a. e^ik is the point on the unit circle at angle k (in radians). The arc distance between them is a-k/n < 1/n

    Edit: Need to clean up algebra. No. It's OK. A little tough to see.
    Last edited by Hartlw; February 8th 2011 at 05:36 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #20
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98
    The purpose of this post is to alert people following this thread of an Edit in previous post and to answer a question of Defunkt's.


    Quote Originally Posted by FernandoRevilla View Post
    Theorem

    Let \lambda\in\mathbb{R} and consider:

    T_{\lambda}:S^1\rightarrow S^1,\quad T_{\lambda}(\theta)=\theta +2\pi\lambda

    Then, each orbit of T_{\lambda} is dense in S^1 if \lambda is irrational

    Proof

    Let \theta\in S^1 . The points on the orbit of \theta are distinct for if T_{\lambda}(\theta)^n=T_{\lambda}^m(\theta) we would have (m-n)\lambda\in \mathbb{Z}, so that n=m . Any infinite set of points on the circle must have a limit point. Thus, given any \epsilon >0 there must be integers n and m for which |T_{\lambda}^n(\theta)-T_{\lambda}^m(\theta)|<\epsilon . Let k=n-m . Then |T_{\lambda}^k(\theta)-\theta|<\epsilon .

    Now T_{\lambda} preserves lengths in S^1 . Consequently, T_{\lambda}^k maps the arc connecting \theta to T_{\lambda}^k(\theta) to the arc connecting T_{\lambda}^k(\theta) and T_{\lambda}^{2k}(\theta) which has length less than \epsilon . In particular it follows that the points \theta,T_{\lambda}^k(\theta),T_{\lambda}^{2k}(\the  ta),\ldots partition S^1 into arcs of length less than \epsilon . Since \epsilon was arbitrary, this completes the proof.


    Fernando Revilla
    The second underlined sentence does not follow from the first. In other words, the fact that a limit point exists does not prove that every point is a limit point.

    Previous post by Hartlw has correct proof:

    OK: For any angle a and any desired degree of accuracy (couched in terms of epsilon),
    m and n exist st (see Rosenlicht above):
    m/n<=a<(m+1)/n
    m<=na<(m+1)
    k=m

    If e^ik = e^im, then k/n<=a<(k+1)/n
    ie, e^ik differs from e^ia by at most 1/n, n arbitrary..

    e^ia is the point on the unit circle at angle a. e^ik is the point on the unit circle at angle k (in radians). The arc distance between them is a-k/n < 1/n
    Follow Math Help Forum on Facebook and Google+

  6. #21
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Actually, it does follow precisely from the first sentence, by the definition of a limit point.

    If you're honestly not going to try to understand his proof and claim that it's wrong when you see a step which you don't understand, then I don't see any point continuing this discussion. Also, as far as I see, the original question has been answered in Fernando's post (well, for anyone but you, but you're not really trying to understand it, only refute it).
    Follow Math Help Forum on Facebook and Google+

  7. #22
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98
    I agree that the symbology of Fernando's proof is unintelligible to me, which is why I proved it myself in intelligible (my opinion) terms.

    But I did follow it enough tio know that the existence of a limit point on an interval does not prove every point is a limit point, as claimed.

    I also note that agreement or approval does not imply understanding.
    Follow Math Help Forum on Facebook and Google+

  8. #23
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Nowhere did he assert that every point is a limit point.

    And a comment regarding your attitude - when someone tries to help you, and you tell them "no, you're wrong and this is how you do it" then don't expect them to help you again. If you don't understand a part of the proof, some of the ideas or some of the notation used - then say so.
    Telling someone that their proof is wrong when, in fact, it is not, will make them extremely reluctant to help you in the future. Asking them to explain more elaborately is a common thing to do - nobody's going to bite you if you do that. On the other hand, if you simply claim that the proof is false, then you probably will be bitten

    Common courtesy can't ever hurt you.
    Last edited by Defunkt; February 8th 2011 at 07:55 AM. Reason: Comment
    Follow Math Help Forum on Facebook and Google+

  9. #24
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by Hartlw View Post
    I agree that the symbology of Fernando's proof is unintelligible to me,



    Fair enough, but then stop claiming that his proof is wrong only because you don't understand some symbols in it!

    Instead, why won't you nicely ask Fernando to explain "his" symbols or anything else in his proof?

    I think also that the proof in the link I gave in my message, with the interval [0,1) modulo 1 instead of S^1

    is pretty straightforward. Did you give it a look and a thought?

    Tonio



    which is why I proved it myself in intelligible (my opinion) terms.

    But I did follow it enough tio know that the existence of a limit point on an interval does not prove every point is a limit point, as claimed.

    I also note that agreement or approval does not imply understanding.
    .
    Follow Math Help Forum on Facebook and Google+

  10. #25
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by Hartlw View Post
    The purpose of this post is to alert people following this thread of an Edit in previous post and to answer a question of Defunkt's.




    The second underlined sentence does not follow from the first.


    Of course it does: it's been shown that \{T^k_\lambda(\theta)\} is infinite, so it has (by Bolzano-Weierstrass,

    since S^1 is compact) a limit point (i.e., a partial limit), which means that some infinite subsquence of \{T^k_\lambda(\theta)\}

    converges to some point in S^1 itself, which means that thus this subsequence is Cauchy and

    that's what's written there...



    In other words, the fact that a limit point exists does not prove that every point is a limit point.


    Why do you think the first part is the same as the above? Where was implied, let alone said, that "every point is a limit point"?

    Tonio



    Previous post by Hartlw has correct proof:

    OK: For any angle a and any desired degree of accuracy (couched in terms of epsilon),
    m and n exist st (see Rosenlicht above):
    m/n<=a<(m+1)/n
    m<=na<(m+1)
    k=m

    If e^ik = e^im, then k/n<=a<(k+1)/n
    ie, e^ik differs from e^ia by at most 1/n, n arbitrary..

    e^ia is the point on the unit circle at angle a. e^ik is the point on the unit circle at angle k (in radians). The arc distance between them is a-k/n < 1/n
    .
    Follow Math Help Forum on Facebook and Google+

  11. #26
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98
    Defunkt is right. I do not understand Fernando's proof. As far as I can tell, it assumes what we are trying to prove. It does not prove that every point of e^ik is a limit point. Furthermore, the existence of a limit point, which Bolzano Weirstrass shows, does not prove that every point of e^ik is a limit point.

    The problem is straight-forward. I do not consider an obscure theorem in obscure symbolism "help." I consider that a game. By the way, I couldn't solve it.

    If someone wants to help, this is what I would cosider help:

    Given any point at angle alpha <= 2pi on the unit circle. What are k and n st:

    k-n2pi-alpha < epsilon

    An intelligible algebraic solution is sought. I'll start it as a thread since this one is hopelessly cluttered, unless some one gives an intelligible answer.

    EDIT: By the way, I assume this is the Jacobi theorem Fernando is referring to, in case anyone is interested.
    Elements of the theory of elliptic ... - Google Books
    Last edited by Hartlw; February 9th 2011 at 04:51 AM.
    Follow Math Help Forum on Facebook and Google+

  12. #27
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    How can you say in one breath that:

    1) You don't understand Fernando's proof
    2) Even though you don't understand it, it has to be wrong

    ????

    The topic creator asked for a proof of his proposition. A proof was given by both tonio and Fernando, and you're the only one who complained about the clarity of the proofs, but instead of asking for help understanding them - you're simply claiming that they're wrong, without giving any argument as to why they are! You're not even trying to read our elaborations which are trying to explain to you why the proofs are correct.

    The problem is straight-forward. I do not consider an obscure theorem in obscure symbolism "help." I consider that a game. By the way, I couldn't solve it.
    You are thinking that because this problem can be easily stated, it can be easily proven. This is a common misconception - for example, see Fermat's Last Theorem - which was conjectured by Fermat on 1637, and it took many generations (more than 350 years) and hundreds, if not thousands of mathematicians to solve - and it can be easily stated: if n>2 is an integer, then there are no integers a,b,c such that a^n + b^n = c^n. It takes zero mathematical knowledge to understand what this conjecture tries to say, but it took the development of an extremely rich and sophisticated field of mathematics to solve it.

    Also, if you'd cared to actually ask us what the notation in Fernando's proof means and how the proof works, we would have gladly elaborated. But I guess you're not even reading anybody's posts but your own, so have fun with that.

    I will ask for a moderator to lock this topic (if the topic creator agrees, of course).
    Follow Math Help Forum on Facebook and Google+

  13. #28
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by Hartlw View Post
    Defunkt is right. I do not understand Fernando's proof. As far as I can tell, it assumes what we are trying to prove.

    No, it doesn't


    It does not prove that every point of e^ik is a limit point.


    Why do you think this is what was to be proven? And "a limit point"...of what??


    Furthermore, the existence of a limit point, which Bolzano Weirstrass shows, does not prove that every point of e^ik is a limit point.


    It also does not prove that Saturn has more than 15 satellites...I think you didn't even understand what was to be proved.


    The problem is straight-forward. I do not consider an obscure theorem in obscure symbolism "help."


    The problem was obviously not that straightforward for you as you apparently
    didn't even understand what had to be proved.
    Certainly that Jacobi's theorem may be not that well-known, but its symbolism is completely
    standard and its difficulty rather medium/low for anyone with 2-3 years of undergraduate studies in maths.



    I consider that a game.


    Much of mathematics is a game: a rather wonderful, intelectually challenging and

    mind-absorbing game. What with this?



    By the way, I couldn't solve it.


    After you belittled someone who tried to help and in the same instance you

    claimed you did prove it? Well, go figure...



    If someone wants to help, this is what I would cosider help:

    Given any point at angle alpha <= 2pi on the unit circle. What are k and n st:

    k-n2pi-alpha < epsilon


    You surely meant: "Given any \epsilon>0....etc", and this is exactly what the proof

    presented by Fernando did



    An intelligible algebraic solution is sought.


    It has been given already, both by Fernando and by myself, at least. That you didn't understand

    them is , perhaps, deplorable and sad, but you could have asked for further explanations.
    Instead, you chose to behave in an even more deplorable way.

    Tonio



    I'll start it as a thread since this one is hopelessly cluttered.
    .
    Follow Math Help Forum on Facebook and Google+

  14. #29
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98
    Quote Originally Posted by Defunkt View Post
    Also, if you'd cared to actually ask us what the notation in Fernando's proof means and how the proof works, we would have gladly elaborated. But I guess you're not even reading anybody's posts but your own, so have fun with that.

    I will ask for a moderator to lock this topic (if the topic creator agrees, of course).
    I note that your offer to explain Fernando's proof is followed immediateley by the suggestion the thread be locked. You mean no one has picked up that I would like an (intelligible) explanation of Fernando's proof? OK, I will make it official. Please explain Fernando's proof in intelligible terms. In particular, how he shows a limit point exists and why every point of e^ik is a limit point.

    PS You need to show every point of e^ik is a limit point on the unit circle in order to show e^ik is dense. But that was all gone over at the beginning. To do that you have to solve: k-n2pi-alpha < epsilon for n and k. Unbelievably, someone asked me to explain epsilon.
    Follow Math Help Forum on Facebook and Google+

  15. #30
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    The suggestion that the thread be locked was because you were not reading the posts explaining the proofs.

    Here is a detailed explanation of Fernando's proof. If there is something you do not understand, say exactly what is not clear.

    Let \lambda \in \mathbb{R} be an arbitrary irrational number, and let T_{\lambda} : S^1 \to S^1 be a function which maps an element of the unit circle to an element of the unit circle.
    Define the mapping by T_{\lambda} (e^{i \theta}) = e^{i ( \theta + 2 \pi \lambda ) } and note that, since each element of S^1 is uniquely determined by its angle \theta with the positive x axis, this mapping is the same as the mapping that Fernando has defined.

    Note that we will use the notation T_{\lambda}^k(e^{i \theta}) to abbreviate \left( T_{\lambda}(e^{i \theta}) \right)^k = =  T_{\lambda}(e^{i \theta}) \circ T_{\lambda}(e^{i \theta}) \circ \overbrace{\ldots}^{\text{k times}} \circ T_{\lambda}(e^{i \theta})

    First, we will show that if m,n are distinct integers ( m \ne n), then T_{\lambda}^m(\theta) - T_{\lambda}^n(\theta) \ne 0 \text{ or } 2 \pi k, \ k \in \mathbb{N} :
    T_{\lambda}^m(\theta) - T_{\lambda}^n(\theta) = (\theta + 2 \pi m \lambda) - (\theta + 2 \pi n \lambda) = \theta - \theta + 2 \pi (m - n) \lambda = 2 \pi (m-n) \lambda
    Now, that equals 0 or  2 \pi k iff m=n or m - n is irrational, but we chose them so that  m \ne n and m - n \in \mathbb{N}, and so T_{\lambda}^m(\theta) - T_{\lambda}^n(\theta) \ne 0 \text{ or }  2 \pi k, \ k \in \mathbb{N}, \ \forall m \ne n \in \mathbb{N}.

    Now, this means that all the elements of the sequence \{T_{\lambda}^k(\theta)\}_{k=1}^{\infty} are distinct, and therefore this sequence has a limit point (by sequential compactness), ie. a convergent subsequence.

    Since that subsequence is cauchy, for any \epsilon>0 there exist integers m<n such that |T_{\lambda}^n(\theta) - T_{\lambda}^m(\theta)| < \epsilon, but then note that T_{\lambda}^n(\theta) - T_{\lambda}^m(\theta) = \theta + 2 \pi n \lambda - \theta - 2 \pi m \lambda = 2 \pi (n-m) \lambda = T_{\lambda}^{n-m}(\theta) - \theta, so let k = n-m to have |T_{\lambda}^k(\theta) - \theta| < \epsilon.

    Now, note that T_{\lambda} is a length-preserving map (ie. it maps an interval of length t to an interval of length t).

    Also, note that T_{\lambda}^k maps the arc connecting the point \theta with T_{\lambda}^k(\theta) to the arc connecting T_{\lambda}^{k}(\theta) with T_{\lambda}^{2k}(\theta).
    Consequently, since |T_{\lambda}^k(\theta) - \theta| < \epsilon and T_{\lambda} preserves lengths, we have that |T_{\lambda}^{2k}(\theta) - T_{\lambda}^k(\theta)| < \epsilon, and by the same fashion |T_{\lambda}^{mk}(\theta) - T_{\lambda}^{(m-1)k}(\theta)| < \epsilon for any m \in \mathbb{N}.

    Now, since all elements of \{T_{\lambda}^k(\theta)\}_{k=1}^{\infty} are distinct, we have that for every \theta \in S^1 there exists an integer n \in \mathbb{N}, such that |\theta - T_{\lambda}^{nk}(\theta)| < \epsilon, and so we have that \{T_{\lambda}^{nk}(\theta)\}_{n=1}^{\infty} is dense in S^1, and in particular any set containing it is dense in S^1 - and such is \{T_{\lambda}^m(\theta)\}_{m=1}^{\infty}.

    Now, if we take \lambda = \frac{1}{2 \pi} (which is clearly irrational) and take \theta = 0, we get that \{T_{\lambda}^m(\theta)\}_{m=1}^{\infty} = \{e^{im}\}_{m=1}^{\infty} is dense in S^1, as conjectured.
    Follow Math Help Forum on Facebook and Google+

Page 2 of 3 FirstFirst 123 LastLast

Similar Math Help Forum Discussions

  1. Unit circle help
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 17th 2011, 09:38 AM
  2. Unit Circle Help
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: October 1st 2009, 10:40 PM
  3. Unit circle
    Posted in the Algebra Forum
    Replies: 3
    Last Post: September 30th 2009, 03:12 AM
  4. Unit Circle Help
    Posted in the Geometry Forum
    Replies: 1
    Last Post: March 23rd 2009, 11:24 PM
  5. Unit Circle-Again
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: November 3rd 2008, 03:23 PM

Search Tags


/mathhelpforum @mathhelpforum