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Math Help - Uniform convergence and differentiation

  1. #1
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    Uniform convergence and differentiation

    (1)

    (a) Let h_n(x)=\frac{sin(nx)}{n}. Show that h_n \rightarrow 0 uniformly on \mathbb{R}. At which points does the sequence of derivatives h'_n converge?

    (b) Modify this example to show that it is possible for a sequence (f_n) to converge uniformly but for (f'_n) to be unbounded.

    (2) Consider the sequence of functions defined by g_n(x)= \frac{x^n}{n}.

    (a) Show (g_n) converges uniformly on [0,1] and find g=lim g_n. Show that g is differentiable and compute g'(x) for all x \in [0,1].

    (b) Now show that (g'_n) converges on [0,1]. Is the convergence uniform? Set h = lim g'_n and compare h and g'. Are they the same?

    Thanks for any and all help!
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  2. #2
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    Show what you did. You asked several questions on convergence of series of functions. You surly had to learn something. Post what you did.
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  3. #3
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    Quote Originally Posted by shadow_2145 View Post
    (1)

    (a) Let h_n(x)=\frac{sin(nx)}{n}. Show that h_n \rightarrow 0 uniformly on \mathbb{R}. At which points does the sequence of derivatives h'_n converge?
    Okay, I will give this a shot.

    Show that h_n \rightarrow 0 uniformly on \mathbb{R}
    |h_n(x)|< 1/n \rightarrow 0
    h_1 = sinx
    h_2 = \frac{sin(2x)}{2}

    Show \forall \epsilon > 0 \exists N_{\epsilon} : n>N \Rightarrow |\frac{sin(nx)}{n} - 0| < \epsilon for all x's in \mathbb{R}

    |\frac{sin(nx)}{n}| \leq 1/n

    So we need to make 1/n < \epsilon \Rightarrow n > \frac{1}{\epsilon}

    (b) Not exactly sure on this part but..

    h'_n(x) = \frac{1}{n}n cos(nx)
    h'_n(\frac{\pi}{2}) = cos(\frac{n\pi}{2}) = 0
    h'_n(\frac{\pi}{4}) = cos(n\cdot\frac{\pi}{4})

    Please let me know how I can improve on this. Thanks!
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  4. #4
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    Quote Originally Posted by shadow_2145 View Post

    (b) Not exactly sure on this part but..

    h'_n(x) = \frac{1}{n}n cos(nx)
    h'_n(\frac{\pi}{2}) = cos(\frac{n\pi}{2}) = 0
    h'_n(\frac{\pi}{4}) = cos(n\cdot\frac{\pi}{4})

    Please let me know how I can improve on this. Thanks!
    h'_n(x) = \sin (nx) while h(x) = 0 \implies h'(x) = 0. The problem is asking to find all x such that \lim h_n'(x) exists and is equal to h'(x), i.e. \lim \sin (nx) = 0. If x \not = 2\pi k then \sin (nx) \not = 0 and so \{ \sin (nx) \} will oscillate. Which means there can be no limit. It seems that x=2\pi k are the points where the converge works.
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