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Math Help - A question about y=sin(n)

  1. #1
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    A question about y=sin(n)

    y=sin(n)     n \in \mathbb{N_+}  ,then y can reach any number in [-1,1]
    may i have a proof?
    i just be very curious why it is ture and it's not my exam ,if the proof is too long ,please give me a e-mail ,thanks
    Last edited by nicklus; May 15th 2010 at 01:22 AM.
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  2. #2
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    Quote Originally Posted by nicklus View Post
    y=sin(n) where n belong to N+ ,then y can reach any number in [-1,1]
    First one should specify what "reach" means: there are (many) real number x\in[-1,1] such that x\neq \sin n for all n\in\mathbb{N} (just because [-1,1] is uncountable).

    However, it is true that any number x\in [-1,1] can be approximated by terms of the sequence (\sin n)_n with arbitrarily good accuracy: there exists a sequence n_k\to +\infty such that \sin n_k\to x. In other words, the set \{\sin n : n\in\mathbb{N}\} is dense in [0,1].

    I think the best way to prove this is to prove a (very slightly) stronger statement about complex numbers: the sequence (e^{in})_{n\in\mathbb{N}} is dense in the complex unit circle. The result follows by projection (if e^{in} is close to e^{i\theta} then \sin n is close to \sin \theta, and we take \theta=\arcsin x).

    The density of \{e^{in} : n\in\mathbb{N}\} in the unit circle comes from a pigeon-hole principle and from the irrationality of \pi.

    The pigeon-hole principle argument is equivalent to the application given in this wikipedia page (paragraph beginning with "A notable problem...") where a=\frac{1}{2\pi}. I guess you can find other references with the keywords I gave you.

    You can find plenty of references about the irrationality of \pi on the web, or just admit it.
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  3. #3
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    Quote Originally Posted by Laurent View Post
    First one should specify what "reach" means: there are (many) real number x\in[-1,1] such that x\neq \sin n for all n\in\mathbb{N} (just because [-1,1] is uncountable).

    However, it is true that any number x\in [-1,1] can be approximated by terms of the sequence (\sin n)_n with arbitrarily good accuracy: there exists a sequence n_k\to +\infty such that \sin n_k\to x. In other words, the set \{\sin n : n\in\mathbb{N}\} is dense in [0,1].

    I think the best way to prove this is to prove a (very slightly) stronger statement about complex numbers: the sequence (e^{in})_{n\in\mathbb{N}} is dense in the complex unit circle. The result follows by projection (if e^{in} is close to e^{i\theta} then \sin n is close to \sin \theta, and we take \theta=\arcsin x).

    The density of \{e^{in} : n\in\mathbb{N}\} in the unit circle comes from a pigeon-hole principle and from the irrationality of \pi.

    The pigeon-hole principle argument is equivalent to the application given in this wikipedia page (paragraph beginning with "A notable problem...") where a=\frac{1}{2\pi}. I guess you can find other references with the keywords I gave you.

    You can find plenty of references about the irrationality of \pi on the web, or just admit it.
    first,thanks to your idea, the idea is on the point ,because that cycle is necessary . And it's not complete, i have learned the complex analysis,the thinking from the friend above is good ,but i want a complete proof, the question refer to much kownledge in math ,esepcailly number theory and complex analysis.only the knowledge on <br />
\pi<br />
may be not enough
    could some give a proof? thanks to the friend above and your idea.
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