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Supremum of sin n

  1. Apr 27, 2005 #1
    I've got a question in analysis:
    How to calculate the supremum of sin n for positive integers n?

    I have tried hard but still cannot figure out it.
    Thanks very much to answer my question in advance! :smile:
     
  2. jcsd
  3. Apr 27, 2005 #2
    This is a question with a fairly involved answer. The answer is 1 (as you might expect), but this is complicated by the fact that [itex]\sin n[/itex] never actually takes the value 1 for any integer [itex]n[/itex]. In fact, [itex]\sin n[/itex] takes infinitely many values in every subinterval of [itex]\left[ -1, 1\right][/itex], but it still manages to "miss" most of them (this is in the same way that there are infinitely many rationals on every real subinterval, but the rationals still have measure 0). In other words, every [itex]x \in \left[ -1, \ 1 \right][/itex] is an accumulation point of [itex]\{\sin n\}_{n=0}^\infty[/itex].

    Try approaching it this way: write [itex]n[/itex] in the form

    [tex]n = 2\pi k + r, \ 0 < r < 2 \pi, \ k \in \mathbb{Z}.[/tex]

    You should prove that this decomposition is unique, in the sense that if you also have [itex]n = 2\pi p + q, \ 0 < q < 2\pi, \ p \in \mathbb{Z}[/itex] then [itex]q=r, \ p=k[/itex], and you should prove that you can always do this. Then, see if you can find a way to show that [itex]r[/itex] can be made arbitrarily close to [itex]\pi / 2 [/itex] by choosing [itex]n[/itex] appropriately, and thus that [itex]\sin n[/itex] can be made arbitrarily close to [itex]\sin (2\pi + \pi/2) = 1[/itex].
     
    Last edited: Apr 27, 2005
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