Results 1 to 5 of 5

Math Help - The Clock Problem

  1. #1
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    The Clock Problem

    I have a blank clock face and a single hand which moves in jumps of one radian. The hand starts at angle 0. For an arbitrary line scratched on the surface at angle alpha, and starting from 0, will the hand ever come within a small angle epsilon of the line? After how many jumps?

    The answer is yes. Please provide a clear proof.
    Last edited by mr fantastic; February 11th 2011 at 12:16 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98
    Quote Originally Posted by Hartlw View Post
    I have a blank clock face and a single hand which moves in jumps of one radian. The hand starts at angle 0. For an arbitrary line scratched on the surface at angle alpha, and starting from 0, will the hand ever come within a small angle epsilon of the line? After how many jumps?

    The answer is yes. Please provide a proof in intelligible terms.

    EDIT: By intelligible I mean to someone who has taken math through calculus.
    Finally got it, what a lulu.

    Use the principle of Archimedes (Ref Rosenlicht or Shilov):

    For any real a and integer N, K exists st

    K \leq Na< K+1, or

    0 \leq a - K/N < 1/N

    So we can come to within 1/N of a after K steps.

    Mathematically, this also proves that e^iK is dense on the unit circle.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98
    Quote Originally Posted by Hartlw View Post
    Finally got it, what a lulu.

    Use the principle of Archimedes (Ref Rosenlicht or Shilov):

    For any real a and integer N, K exists st

    K \leq Na< K+1, or

    0 \leq a - K/N < 1/N

    So we can come to within 1/N of a after K steps.

    Mathematically, this also proves that e^iK is dense on the unit circle.
    WRONG!!!!! (MADE THE SAME MISTAKE TWICE) All I have done is made the obvious statement that K exists st after K steps of 1/N each I will be within 1/N of a.

    Given K I can write:

    K/2pi = M + b/2pi , 0 <b <2pi , (b can't be zero or 2pi because that would mean pi is rational)

    but given "a" I can't find K so that "b" is within \epsilon of a.

    EDIT Sorry about bumping someone off the top of the line with this. I give up.
    Last edited by Hartlw; February 11th 2011 at 12:45 PM. Reason: b to b/2pi, add "or 2pi"
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1
    Here is an observation that may help.

    Suppose \epsilon > 0 is given. Choose n sufficiently large that 2 \pi / n < \epsilon. Partition the circle into n arcs

    [0, 2 \pi /n), [2 \pi /n, 4 \pi /n), [4 \pi /n, 6 \pi /n), \dots , [2 (n-1) \pi /n, 2 \pi).

    Consider the angles 1, 2, ..., n , n+1 (radians). No two of these angles can be the same (remember that angles that differ by a an integer multiple of 2 \pi are equivalent), because if j and k were equivalent as angles we would have j-k = 2 m \pi for some integer m, contradicting the irrationality of \pi.

    Since there are n arcs and n+1 angles, by the Pigeonhole Principle one of the arcs must contain at least two angles, say j and k. Let's say j < k. Define N = k-j. Then N is an angle less, in absolute value, than the width of the arc, 2 \pi / n, i.e. N = 2 m \pi + \theta, where m is an integer and | \theta | < 2 \pi /n < \epsilon.

    Now consider the multiples of N; what can you say about them as angles?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98
    Quote Originally Posted by Hartlw View Post
    I have a blank clock face and a single hand which moves in jumps of one radian. The hand starts at angle 0. For an arbitrary line scratched on the surface at angle alpha, and starting from 0, will the hand ever come within a small angle epsilon of the line? After how many jumps?
    Each 7 steps of 1 you advance .7168
    7(1) = 2pi + .7168

    Each 9 increments of .7168 you advance .1680
    9(.7168) = 2pi + .1680

    Each 38 increments of .1680 you advance .1008
    38(.1680) = 2pi + .1008

    63(.1008) = 2pi + .0672
    94(.0672) = 2pi + .0336
    187(.0336) = 2pi + .0001

    The general pattern is:
    7/2pi = 1/N1 + r1, N1=1, r1<1
    r1/2pi = 1/N2 + r2, N2>N1, r2<r1
    r2/2pi = 1/N3 + r3, N3>N2, r3<r2
    .........
    any rNi = 0 implies pi rational (substitute and work backwards).

    Ni=>infinity, rNi=>0

    So the circle is divided into aritrarily small increments as the hand steps around and we can come as close as we please to the line scribed on the face.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Clock Problem
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: September 15th 2009, 05:45 PM
  2. clock problem
    Posted in the Algebra Forum
    Replies: 4
    Last Post: August 19th 2009, 10:40 PM
  3. Clock Problem
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: March 4th 2009, 08:12 AM
  4. Clock problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 19th 2008, 11:12 AM
  5. alarm clock problem.
    Posted in the Advanced Applied Math Forum
    Replies: 0
    Last Post: May 2nd 2007, 08:13 PM

Search Tags


/mathhelpforum @mathhelpforum