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Does |Sin(n)| Converge?

  1. Jul 3, 2011 #1
    Hi, All:

    I am trying to see if a_n:={|Sin(n)|}, with n=1,2,... and | . | standard absolute value,
    is convergent. I know the set {k.pi}, k=1,2,... is dense in [0,1] (pi is equidistributed mod1) , and we have that Sin(n)=Sin(n+pi), but it seems like {|Sinn|} is dense in [0,1], so that it cannot have a limit (i.e., a unique limit point). Any Ideas?
  2. jcsd
  3. Jul 4, 2011 #2
    No, it doesn't converge. From the equidistribution theorem, a result very similar to the one you quoted holds: {k/(2pi) mod 1} is dense in [0, 1]. Scaling everything up by 2pi, {k mod 2pi} is dense in [0, 2pi]. Now 0 and pi/2 are both limit points of this set, so they're approached by convergent sequences s1 and s2 respectively. But then the limit of |sin(an)| on these two sequences is |sin(0)|=0 and |sin(pi/2)|=1 respectively, from continuity. The lim sup and lim inf are then unequal, so the original sequence doesn't converge. This same reasoning does show that {|sin(n)|} is dense on [0, 1].
  4. Jul 5, 2011 #3
    Thanks, Josh; a new question came to me:

    Is there a way of determining any actual value assumed by {|sin(n)|}? Or, are we sure there are rationals or irrationals in the sequence? We can of course exclude {|sinx|: x irrational, and values like Sqr2/2 , as the image of Pi/4; do you know of any result, e.g.,
    like with elements of the Cantor set and the base-3 representation (excluding, I think, strings with a 0 in them) about this set?
  5. Jul 5, 2011 #4
    The only relevant result I'm aware of is the http://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem" [Broken] (actually, Lindemann's half is sufficient) which says that e^a where a != 0 is any algebraic number is transcendental. [Briefly, an algebraic number is a root of some polynomial equation with integer coefficients. A transcendental number is not. Transcendental numbers are in particular irrational, since the root of ax - b is b/a.] It turns out that any rational polynomial with integer coefficients evaluated at a transcendental number gives a transcendental result. I don't know a name for this result. The only proof I've seen is my own, since it was an exercise in a Galois theory book of mine.

    In any case, if n is a positive integer, ni is algebraic [(x - ni)(x + ni) = x^2 + n^2], so e^(ni) is transcendental, so

    (e^(in))^2 - 1)/(e^(in))
    = e^(in) - e^(-in)

    is transcendental, so

    1/(2i) (e^(in) - e^(-in))
    = sin(n)

    is transcendental, so |sin(n)| is transcendental. This immediately rules out eg. sqrt(2)/2, since this is obviously algebraic. An arbitrary finite nesting of radicals is also ruled out: sqrt(2 + sqrt(2))/2 is never hit, for instance.

    To be clear the sequence {|sin(n)|} where n is a positive integer is composed only of transcendental numbers, so contains no irrationals. You can actually use the formula listed http://en.wikipedia.org/wiki/List_o....2C_cosine.2C_and_tangent_of_multiple_angles" to write sin(n) in terms of sin(1), cos(1) = sqrt(1 - sin^2(1)), and positive exponents of these two. That is, sin(n) is in the field extension Q(sin(1), cos(1)).
    Last edited by a moderator: May 5, 2017
  6. Jul 6, 2011 #5
    Excellent; very helpful, Josh. Thanks.
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