Results 1 to 2 of 2

Thread: Analysis Help - 1

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    13

    Analysis Help - 1

    IV. 1. Explain why it is that if a series converges (pointwise) on the closed interval [a, b] and converges uniformly on the open interval (a, b), then the series must converge uniformly on the closed interval [a, b]. (Note that to answer this question you must provide a proof that the series converges uniformly at the endpoints of the interval.)


    IV. 2. Give an example of a function series that converges (pointwise) at every point of (a, b), each summand is continuous at every point in [a, b], but the series does not converge at every point of [a, b]. Formally prove that your function series satisfies all these conditions.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by pila0688 View Post
    IV. 1. Explain why it is that if a series converges (pointwise) on the closed interval [a, b] and converges uniformly on the open interval (a, b), then the series must converge uniformly on the closed interval [a, b]. (Note that to answer this question you must provide a proof that the series converges uniformly at the endpoints of the interval.)
    For the series $\displaystyle \{s_n(x)\}$ to converge uniformly to $\displaystyle s(x)$ on $\displaystyle (a,b)$ for all $\displaystyle \varepsilon>0$ there exists an $\displaystyle N_{\varepsilon}$ such that for all $\displaystyle n>N_{\varepsilon}$:

    $\displaystyle |s_n(x)-s(x)|<\varepsilon$

    and $\displaystyle N_{\varepsilon}$ is independednt of $\displaystyle x$.

    As the series converges at $\displaystyle x=s$ and $\displaystyle x=b$, for all $\displaystyle \varepsilon>0$ there exist $\displaystyle N_{1,\varepsilon}$ and $\displaystyle N_{2,\varepsilon}$ such that for all $\displaystyle n>N_{1,\varepsilon}$:

    $\displaystyle |s_n(a)-s(a)|<\varepsilon$

    and for all $\displaystyle n>N_{2,\varepsilon}:$

    $\displaystyle |s_n(b)-s(b)|<\varepsilon.$

    Now for $\displaystyle \varepsilon>0$ let $\displaystyle M_{\varepsilon}=\max(N_{\varepsilon}$, $\displaystyle N_{1,\varepsilon},N_{2,\varepsilon}).$

    Now for all $\displaystyle x$ in $\displaystyle [a,b]$ we have for all $\displaystyle n>M_{\varepsilon}:$

    $\displaystyle |s_n(x)-s(x)|<\varepsilon$

    which proves that $\displaystyle \{ s_n(x) \}$ is uniformly convergent on $\displaystyle [a,b]$

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with Analysis please :)
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Oct 16th 2009, 04:38 AM
  2. analysis help
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Oct 8th 2009, 03:11 PM
  3. analysis help
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Jan 19th 2009, 02:38 PM
  4. analysis
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Dec 15th 2008, 03:07 PM
  5. Analysis Help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Sep 8th 2008, 05:59 PM

Search Tags


/mathhelpforum @mathhelpforum