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Math Help - how to show if they r uni. convergent

  1. #1
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    how to show if they r uni. convergent

     f_{n}(x)=log(x+\frac{1}{n}) <br />
f_{n}(x)=cos(\frac{x}{n})
    on (0,1)
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by silversand View Post
     f_{n}(x)=log(x+\frac{1}{n}) <br />
f_{n}(x)=cos(\frac{x}{n})
    on (0,1)
    I'm a little confused about what you have written.

    I assume those are two different functions?

    f_n(x)=\log(x+\frac{1}{n}) and f_n(x)=\cos(\frac{x}{n})

    Also, are you using \log(x) to mean the same thing as \ln(x)?
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  3. #3
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    yes

    yes. they are two cases. and log means ln
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by silversand View Post
     f_{n}(x)=log(x+\frac{1}{n}) <br />
f_{n}(x)=cos(\frac{x}{n})
    on (0,1)
    Second function: We know that this function converges to \cos(0)=1, the question is whether it does so uniformly. Using the difference-to-product identity, we have:

    \left|\cos(0)-\cos\left(\frac{x}{n}\right)\right| = \left|-2\sin\left(\frac{x}{2n}\right)\sin\left(-\frac{x}{2n}\right)\right|\leq 2\left|\sin\left(\frac{x}{2n}\right)\right| because |\sin(y)|\leq 1 for all y.

    Since 2\sin\left(\frac{x}{2n}\right) is increasing and positive on (0,1), we know that 2\left|\sin\left(\frac{x}{2n}\right)\right|\leq 2\sin\left(\frac{1}{2n}\right).

    It is now clear that for all \epsilon>0, we can choose n large enough so that \left|1-\cos\left(\frac{x}{n}\right)\right| \leq 2\sin\left(\frac{1}{2n}\right)<\epsilon

    If you need an explicit formula, I think choosing n>\frac{1}{2\sin^{-1}(\frac{\epsilon}{2})} will do the trick.
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