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Convergence of sin(x/n) in A[0,pi]

  1. Nov 27, 2007 #1
    1. The problem statement, all variables and given/known data

    Show that the sequence real-valued functions fn(x) = sin(x/n) converges uniformly to f(x) = 0 on A=[0,pi]

    2. Relevant equations

    I don't believe there are any

    3. The attempt at a solution

    Well, my first thought was that if x/n = pi then sin(x/n) is obviously going to be zero. I really can't get past that, I mean it seems like stating that and that if x/n is any multiple of pi then sin(x/n) will be zero. I guess I can't see past the obvious and into how this problem might actually be challenging.
  2. jcsd
  3. Nov 27, 2007 #2


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    You are supposed to show that fn(x) goes to zero as n goes to infinity for *any* value of x between 0 and pi.
  4. Nov 27, 2007 #3
    Oh. Well then. You can't use euler's because the function is real valued, correct? Or, can you expand it and just leave off the imaginary part?

    I can't think of another way to show that this function is going to zero.
  5. Nov 27, 2007 #4


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    Well, as n goes to infinity, x/n goes to zero for all x on [0,pi] And sin(0)=0
  6. Nov 27, 2007 #5
    If the sup norm ( sup |fn(x)-f(x)|, over x in X) goes to zero then f_n converges uniformly to f on X. This is biconditional.

    I suspect you can also show this from the definition of uniform convergence.
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