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Uniform convergence

  1. Oct 6, 2012 #1
    1. The problem statement, all variables and given/known data
    I need to show that [itex]f_{n}[/itex]=sin([itex]\frac{z}{n}[/itex]) converges uniformly to 0.

    2. Relevant equations
    So I need to find K([itex]\epsilon[/itex]) such that [itex]\forall[/itex][itex]n \geq K[/itex]
    |sin([itex]\frac{z}{n}[/itex])|<[itex]\epsilon[/itex]

    I'm trying to prove this in an annulus: [itex]\alpha\leq |z| \leq\beta[/itex]

    3. The attempt at a solution
    I'm having trouble because no matter what I choose for K I can't get the epsilon to come out.
    I'm trying something like K([itex]\epsilon[/itex])=[itex]\frac{1}{\alpha\epsilon}[/itex].

    My problem is that I can't say that sin([itex]\frac{z}{n}[/itex])<sin([itex]\frac{\alpha}{n}[/itex])

    Which is how I've been doing these uniform convergence ones (recasting in terms of [itex]\alpha[/itex] instead of z).

    Anyways I was hoping I could get some help on how to proceed.

    Thanks!
     
    Last edited: Oct 6, 2012
  2. jcsd
  3. Oct 6, 2012 #2

    jbunniii

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    Try using the fact that [itex]|\sin(x)| \leq |x|[/itex] for all [itex]x[/itex].
     
  4. Oct 6, 2012 #3
    Does that work for complex numbers too? It strikes me that if y=Im(z), then sin(z)<=cosh(y) ??

    Writing z as z=x+iy, then sin(z)=[itex]\frac{e^{iz}-e^{-iz}}{2i}[/itex] becomes: [itex]\frac{1}{2i}[/itex]([itex]e^{-y+ix}[/itex]-[itex]e^{y-ix}[/itex]) Then taking the magnitude and using the triangle inequality, it seems it'd come to [itex]\frac{1}{2}[/itex]|[itex]e^{y}[/itex]|+[itex]\frac{1}{2}[/itex]|[itex]e^{-y}[/itex]|

    So then I can't use |sin(z)|[itex]\leq[/itex]|z| ????
     
    Last edited: Oct 7, 2012
  5. Oct 7, 2012 #4
    Any suggestions? Or comments?
     
  6. Oct 7, 2012 #5

    jbunniii

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    Sorry, I didn't notice you were working with complex numbers. In that case if we write [itex]z = x + iy[/itex], then
    [tex]\sin(z) = \sin(x) \cosh(y) + i \cos(x) \sinh(y)[/tex]
    so
    [tex]|\sin(z)| \leq |\sin(x) \cosh(y)| + |\cos(x) \sinh(y)| \leq
    |\cosh(y)| + |\sinh(y)|[/tex]
    Therefore,
    [tex]|\sin(z/n)| \leq |\cosh(y/n)| + |\sinh(y/n)|[/tex]
    So the problem reduces to showing that cosh(y/n) and sinh(y/n) converge uniformly to zero in the annulus as [itex]n \rightarrow \infty[/itex].
     
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