Results 1 to 5 of 5

Math Help - functions

  1. #1
    Member
    Joined
    Oct 2008
    Posts
    156

    functions

    Let  f^{(n)}(x) := 2n when  x \in [1/2n, 1/n] and  f^{(n)}(x) := 0 for all other values of  x . Why does  f^{(n)} converge pointwise to  f(x) := 0 ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by manjohn12 View Post
    Let  f^{(n)}(x) := 2n when  x \in [1/2n, 1/n] and  f^{(n)}(x) := 0 for all other values of  x . Why does  f^{(n)} converge pointwise to  f(x) := 0 ?

    Pointwise convergence depends on both n and x.

    So for any fixed x if it is out side of [0,1]

    f^{n}=0 for all n

    so we are okay.

    Now for any x \in [0,1]

    There exits an N such that for all n> N f^{n}=0

    The Key point is that it depends both on x and n. I hope this helps
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2008
    Posts
    156
    Quote Originally Posted by TheEmptySet View Post
    Pointwise convergence depends on both n and x.

    So for any fixed x if it is out side of [0,1]

    f^{n}=0 for all n

    so we are okay.

    Now for any x \in [0,1]

    There exits an N such that for all n> N f^{n}=0

    The Key point is that it depends both on x and n. I hope this helps
    I think you mean  [1/2, 1] . Also why is this true: There exits an N such that for all n> N f^{n}=0? Because  f^{(100)}(x) := 200 for all  x \in [1/2, 1 ] .
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by manjohn12 View Post
    I think you mean  [1/2, 1] . Also why is this true: There exits an N such that for all n> N f^{n}=0? Because  f^{(100)}(x) := 200 for all  x \in [1/2, 1 ] .
    pay attention to your interval. [1/2,1] is the BIGGEST it will be. for your example, x = 100, we have the interval [1/200, 1/100]. this is why TES considers the interval (0,1], because [1/(2n), 1/n] \to \{ 0 \} (note, 0 is never actually in the interval)

    so as n gets large, the interval for which f^{(n)} \ne 0 gets smaller and smaller...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by manjohn12 View Post
    I think you mean  [1/2, 1] . Also why is this true: There exits an N such that for all n> N f^{n}=0?
    I meant (0,1]

    What we have is a sequnce of intervals.

    When n=1 [\frac{1}{2},1]
    n=2 [\frac{1}{4},\frac{1}{2}]
    n=3 [\frac{1}{6},\frac{1}{3}]
    n=4 [\frac{1}{8},\frac{1}{4}]
    ...

    So eventally every x in (0,1]

    Is not in [\frac{1}{2n},\frac{1}{n}]

    so f^{n}=0
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: April 15th 2010, 05:50 PM
  2. Replies: 3
    Last Post: February 23rd 2010, 04:54 PM
  3. Replies: 11
    Last Post: November 15th 2009, 11:22 AM
  4. Replies: 7
    Last Post: August 12th 2009, 04:41 PM
  5. Replies: 1
    Last Post: April 15th 2008, 09:00 AM

Search Tags


/mathhelpforum @mathhelpforum