Page 1 of 2 12 LastLast
Results 1 to 15 of 20

Math Help - prove sin(x) is uniformly continuous.

  1. #1
    Member
    Joined
    Aug 2009
    Posts
    94

    prove sin(x) is uniformly continuous.

    Prove directly from the definition of uniform continuity that \sin(x) is uniformly continuous on (-\infty,\infty).

    (Hint: a unit circle may help to make an estimation.)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    12
    prove that |\sin x-\sin y|\le|x-y| holds for x,y\in\mathbb R and you're done.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2009
    Posts
    94
    Anything wrong with this argument?

    given \epsilon,\delta > 0
    |\sin(x) - \sin(y)| \leq |x - y| < \epsilon = \delta
    Last edited by cgiulz; October 25th 2009 at 09:32 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Double post.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1

    |m-n|\leq |m|-|n| is wrong. Take m=0, n=1 for example.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Aug 2009
    Posts
    94
    Quote Originally Posted by Bruno J. View Post
    |m-n|\leq |m|-|n| is wrong. Take m=0, n=1 for example.
    Typo; I was trying to apply the triangle inequality. It is not even necessary anyway. Any thoughts on the edited post?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    Since the derivative is bounded, using the mean value theorem you get that the function is uniformly continous.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by cgiulz View Post
    Prove directly from the definition of uniform continuity that \sin(x) is uniformly continuous on (-\infty,\infty).

    (Hint: a unit circle may help to make an estimation.)
    Here are some tools that will help you.

    |\sin x-\sin y|=\left|2\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)\right|

    |\cos\theta|\leq1 and you can use the unit circle to show that |\sin\theta|\leq|\theta|.

    Let \delta=\epsilon and don't look back.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Aug 2009
    Posts
    94
    Quote Originally Posted by Jose27 View Post
    Since the derivative is bounded, using the mean value theorem you get that the function is uniformly continous.
    Yes, but how can it be done directly from the definition?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by Jose27 View Post
    Since the derivative is bounded, using the mean value theorem you get that the function is uniformly continous.
    The only thing is that this requires him to assume \sin x is differentiable and furthermore that it's derivative is \cos x. (If they haven't proved that \sin x is continuous yet, it's unlikely they've proved its derivative is \cos x.)
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Aug 2009
    Posts
    94
    Quote Originally Posted by redsoxfan325 View Post
    The only thing is that this requires him to assume \sin x is differentiable and furthermore that it's derivative is \cos x. (If they haven't proved that \sin x is continuous yet, it's unlikely they've proved its derivative is \cos x.)
    This is true.

    Go yanks
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by cgiulz View Post
    Go yanks


    I hope the Phillies can pull it off two years in a row.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    You could say that \sin x is unif. cont. in [0,2 \pi] and since it's periodic it's unif. cont. in \mathbb{R}

    Edit. Apparently I keep avoiding the definition of uniform cont.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member
    Joined
    Aug 2009
    Posts
    94
    Quote Originally Posted by cgiulz View Post
    Anything wrong with this argument?

    given \epsilon,\delta > 0
    |\sin(x) - \sin(y)| \leq |x - y| < \epsilon = \delta
    I think this is correct, is it not?
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    That is true, yes, but I think you need to show that |\sin x-\sin y|\leq|x-y|, not just state it.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: April 18th 2011, 08:19 AM
  2. Replies: 3
    Last Post: April 18th 2011, 07:24 AM
  3. prove that the composition gof:A-> C is uniformly continuous
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 9th 2011, 12:49 PM
  4. how to prove this function to be uniformly continuous ?
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: March 13th 2011, 07:56 AM
  5. prove that f is uniformly continuous..
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 25th 2010, 09:15 AM

Search Tags


/mathhelpforum @mathhelpforum