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Math Help - Uniform Convergence of Series

  1. #1
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    Uniform Convergence of Series

    Hi,

    Show that the series  \sum \frac{x^n}{1+x^n} converges uniformly on [0,a] for any a in (0,1)

    There is a theorem in the book that says that is the series of functions satisfies the cauchy criterion uniformly on a set S, then the series converges uniformly on S. So I guess i'm supposed to use that. I'm not sure how to do a cauchy criterion proof though.

    Thanks
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    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by tbyou87 View Post
    Hi,

    Show that the series  \sum \frac{x^n}{1+x^n} converges uniformly on [0,a] for any a in (0,1)

    There is a theorem in the book that says that is the series of functions satisfies the cauchy criterion uniformly on a set S, then the series converges uniformly on S. So I guess i'm supposed to use that. I'm not sure how to do a cauchy criterion proof though.

    Thanks
    we haven't discussed this yet (it would be this coming semester) but im reading in advance.. would Weierstrass M-Test help?
    thanks..
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tbyou87 View Post
    Hi,

    Show that the series  \sum \frac{x^n}{1+x^n} converges uniformly on [0,a] for any a in (0,1)

    There is a theorem in the book that says that is the series of functions satisfies the cauchy criterion uniformly on a set S, then the series converges uniformly on S. So I guess i'm supposed to use that. I'm not sure how to do a cauchy criterion proof though.

    Thanks
    Quote Originally Posted by kalagota View Post
    we haven't discussed this yet (it would be this coming semester) but im reading in advance.. would Weierstrass M-Test help?
    thanks..
    i was thinking about the Weierstraβ M-Test, but i cannot think of a nice upper bound.

    use the theorem that says: If the series \sum g_n converges uniformly on a set S, then \lim_{n \to \infty} \left[  \sup \{ |g_n(x)| : x \in S \}\right] = 0

    and use the contrapositive


    TPH should be able to come up with a more elegant solution though, i'm sure he can use the Cauchy criterion proof/disproof
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    Doesn't that theorem only apply to sequences and not series?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tbyou87 View Post
    Doesn't that theorem only apply to sequences and not series?
    it is somewhat different from the series one, and as you can see, the theorem is talking about a series. it turns out that the series wont uniformly converge if the sequence of functions it is summing does not. the theorem was proven in your book as an example, look at it
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    Quote Originally Posted by Jhevon View Post
    Weierstraβ M-Test
    You started doing German "S" too?

    TPH should be able to come up with a more elegant solution though, i'm sure he can use the Cauchy criterion proof/disproof
    0\leq \frac{x^n}{x^n+1} \leq \frac{x^n}{1} = x^n \leq a^n.
    But \sum_{n=0}^{\infty}a^n < \infty since a\in (0,1).

    Note: Uniform convergence might not work on the full interval [0,1].
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    You started doing German "S" too?
    hehe, yes, and it is for "ss"

    0\leq \frac{x^n}{x^n+1} \leq \frac{x^n}{1} = x^n \leq a^n.
    But \sum_{n=0}^{\infty}a^n < \infty since a\in (0,1).

    Note: Uniform convergence might not work on the full interval [0,1].
    how would you show that it's not, without the limit theorem that i posted earlier?
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    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Jhevon View Post
    i was thinking about the Weierstraβ M-Test, but i cannot think of a nice upper bound.
    i was thinking if 1 can be an upper bound.. oh well, a^n is..
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kalagota View Post
    i was thinking if 1 can be an upper bound.. oh well, a^n is..
    the series \sum_{n = 0}^{\infty} 1 does not converge, so using that makes no sense. TPH addressed this already, so no worries
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    There is a follow up question that is similar.

    Does the series converge uniformly on [0,1)? Prove your idea.

    Wouldn't that just be the same proof that TPH did using the weierstrass M-test or am I missing something.

    Thanks
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    Quote Originally Posted by Jhevon View Post
    hehe, yes, and it is for "ss"

    how would you show that it's not, without the limit theorem that i posted earlier?
    First I wanted to say [0,1) not [0,1] because it does not converge at 1. But it does converge (in fact uniformly as shown) on [0,1).

    1)The functions f_n(x) = \frac{x^n}{x^n+1} are uniformly continous functions on [0,1).

    2)Therefore the series of functions g(x) = \sum_{n=0}^{\infty}f_n(x) is uniformly continous on [0,1).*

    3)Thus, if x_n is Cauchy in [0,1) then g(x_n) is also Cauchy (properties of uniformly continous functions). But as you can see as you get closer to 1 the value of g(x) becomes unbounded. So it cannot be Cauchy.

    4)Thus, our initial assumption that the convergence was uniform on the full interval [0,1) was wrong.

    *)Because uniformly continous functions that converge uniformly will convergence to a uniformly continous functions.
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    Quote Originally Posted by ThePerfectHacker View Post
    First I wanted to say [0,1) not [0,1] because it does not converge at 1. But it does converge (in fact uniformly as shown) on [0,1).

    4)Thus, our initial assumption that the convergence was uniform on the full interval [0,1) was wrong.
    Sorry I'm a little bit confused. Does the series converge uniformly on [0,1) or not?

    Thanks again
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    Quote Originally Posted by tbyou87 View Post
    Sorry I'm a little bit confused. Does the series converge uniformly on [0,1) or not?
    No. (That was not part of your original problem, that was something I added.)
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