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Thread: Limit of sin(2^n) as n-> inifinity

  1. #1
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    Post Limit of sin(2^n) as n-> inifinity

    I want to calculate this limit for quite some days now and I have made almost no progress whatsoever.
    (limit of sin(2^n) as n-> infinity, neN ) I suspect either lim(sin(2^n)) = 0 or it is non-existing.
    Thus fur I have used numerous trigonometry properties resulting to nowhere,
    also I tried to prove it converges to 0 but turned out my calculations lead it diverging which is impossible.
    I can't seem to find a proper way of calculating/solving this other than an informal theoretical way.

  2. #2
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    Quote Originally Posted by Kiyo View Post
    I want to calculate this limit for quite some days now and I have made almost no progress whatsoever.
    (limit of sin(2^n) as n-> infinity, neN ) I suspect either lim(sin(2^n)) = 0 or it is non-existing.
    Thus fur I have used numerous trigonometry properties resulting to nowhere,
    also I tried to prove it converges to 0 but turned out my calculations lead it diverging which is impossible.
    I can't seem to find a proper way of calculating/solving this other than an informal theoretical way.
    Why that is not acceptable?

    How do you calculate limit of an oscillating function?
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #3
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    I don't seem to know how to calculate a limit of an oscillating function.
    When I said an informal theoretical way I meant a poor explanation, where as in physics an oscillating object, take a string for example, if it oscillates with an incredible amount of angular velocity, the width it has is negligible, thus close to 0.
    What I tried to do was prove somehow that |sin(2^n)| is a decreasing function.
    Correct me if I'm wrong about this.

  4. #4
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    Quote Originally Posted by Kiyo View Post
    I don't seem to know how to calculate a limit of an oscillating function.
    When I said an informal theoretical way I meant a poor explanation, where as in physics an oscillating object, take a string for example, if it oscillates with an incredible amount of angular velocity, the width it has is negligible, thus close to 0.
    What I tried to do was prove somehow that |sin(2^n)| is a decreasing function. ..... No - value of sin(2^n) will depend on the remainder of (2^n)/(2π).
    Correct me if I'm wrong about this.
    |sin(2^n)| ≤ 1
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  5. #5
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    How would you calculate:

    [tex]\displaystyle{\lim_{x \to \infty} \sin(x)}[/tex]
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  6. #6
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    I would say since -1≤ sin(x) ≤ 1 and x∈R
    sin(x) oscillates between all the values within -1,1 thus the


    lim(sin(x)) as x->infinity

    Does not exist.
    But I thought otherwise about sin(2^n) as n
    ∈N,
    it doesn't take all the values between -1,1 and 2^n isn't a linear function.


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