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Math Help - Sequence of functions-convergence

  1. #1
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    Sequence of functions-convergence

    Let the sequence of functions
    gn : R----R be given by g _{n}=x/(1+nx^{2})

    (b) Show that gn converges to g uniformly.
    Do i need to show g_{n} converge to g=x/(1+x^{2})as n tends to infinity?

    How do i show g_{n} converges to g uniformly.Thanks
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by charikaar View Post
    Let the sequence of functions
    gn : R----R be given by g _{n}=x/(1+nx^{2})

    (b) Show that gn converges to g uniformly.
    Do i need to show g_{n} converge to g=x/(1+x^{2})as n tends to infinity?

    How do i show g_{n} converges to g uniformly.Thanks
    Show that g_n(x)\to g(x) in the supremum norm metric. Are you familiar with these terms?
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    From part a) g(x)= lim (g_{n}) as n goes to infinity.

    I worked out g(x)=0 how do i show g_{n} converges to zero uniformly.
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    Quote Originally Posted by charikaar View Post
    Let the sequence of functions
    gn : R----R be given by g _{n}=x/(1+nx^{2})

    (b) Show that gn converges to g uniformly.
    Do i need to show g_{n} converge to g=x/(1+x^{2})as n tends to infinity?

    How do i show g_{n} converges to g uniformly.Thanks

    Wouldn't g have to be 0 since as n approaches infinity, x/(1+nx^{2}) approaches 0 for each x?
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    yes g=0.

    if given epsion>0, how do i choose an integer N to satisfy uniform convergence condtion.

    thanks
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by charikaar View Post
    From part a) g(x)= lim (g_{n}) as n goes to infinity.

    I worked out g(x)=0 how do i show g_{n} converges to zero uniformly.
    Exactly, we claim that \frac{x}{1+nx^2}=f_n(x)\overset{\text{unif.}}{\lon  grightarrow}0.

    To see this we need only prove that for every \varepsilon>0 there exists some N\in\mathbb{N} such that N\leqslant n\implies \left\|\frac{x}{1+nx^2}\right\|_{\infty}<\varepsil  on.

    But, notice that \frac{x}{1+nx^2}\leqslant\frac{1}{2\sqrt{n}}. And so, \left\|\frac{x}{1+nx^2}\right\|_{\infty}\leqslant\  frac{1}{2\sqrt{n}}.

    It follows that choosing n\geqslant\frac{1}{4\varepsilon^2} ensures that \left\|\frac{x}{1+nx^2}\right\|_{\infty}<\varepsil  on.

    The conclusion follows.
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    Quote Originally Posted by charikaar View Post
    Let the sequence of functions
    gn : R----R be given by g _{n}=x/(1+nx^{2})

    (b) Show that gn converges to g uniformly.
    Do i need to show g_{n} converge to g=x/(1+x^{2})as n tends to infinity?

    How do i show g_{n} converges to g uniformly.Thanks
    The problem has two parts:

    PART A:

    Here we must show that the sequence of functions converges to a function f
    Since for each xεR and for each ε>0

    |\frac{x}{1+nx^2}|= \frac{|x|}{1+nx^2}\leq\frac{1}{n|x|} and by choosing \kappa>\frac{1}{\epsilon|x|},then

    for all ,n: n\geq\kappa\Longrightarrow|\frac{x}{1+nx^2}|<\epsi  lon
    we can conlude that the limit is the zero function.

    PART B:

    Here we must show that the sequence converges uniformly to the zero function:

    This can be done by many ways.One of them is by appling the definition of uniform continuity i.e:

    given an ε>0 we must find a natural No κ such that:

    for all ,n,x:  n\geq\kappa and x\in R\Longrightarrow|\frac{x}{1+nx^2}|<\epsilon.

    The proof of that is based on the following inequality:

    (\sqrt n|x| - 1)^2)\geq 0 for each real,x and natural,n.

    Then we can easily deduce from that inequality that:

    \frac{|x|}{1+nx^2}\leq\frac{1}{2\sqrt n} ,And since

    |\frac{x}{1+nx^2}|= \frac{|x|}{1+nx^2} if we choose \kappa\geq\frac{1}{4\epsilon^2} ,then:


    for all ,n,x:  n\geq\kappa and x\in  R\Longrightarrow|\frac{x}{1+nx^2}|<\epsilon.

    Thus the sequence converges uniformly to the zero function.

    Another way is to use the bounded sequence of functions theorem .

    And since the sequence of functions is bounded ,since for all ,n and,x \frac{x}{1+nx^2}\leq\frac{1}{2\sqrt n}

    because (\sqrt n.x - 1)^2\geq 0.and also the zero function is also bounded then according to the theorem the sequence of functions converges uniformly to the zero function
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by xalk View Post
    The problem has two parts:
    Really, not to be combative but what is the point of this post? I often post after a question has been answered if I have something new to add, but you literally took the same argument I made and just did it in full. Why?
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    Quote Originally Posted by Drexel28 View Post
    but you literally took the same argument I made and just did it in full. Why?
    Which is your argument that is the same with my argument??

    My proofs are based on the definition of uniform continuity and on the bounded theorem .

    Your proof i do not know where it is based.
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