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  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Series of function...

    Let, f_n(x)=xarctg(nx) when 0\leq x<\infty.

    I need to find:
    1. Limit function (Which I found: \frac{\pi}{2}x )

    2. Checking the unitary convergence. (So, here is my problem)


    Thank you all!
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  2. #2
    Senior Member roninpro's Avatar
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    Be careful! The limit function is not exactly \frac{\pi}{2}x. Make sure you consider two cases: x\geq 0 and x< 0.

    Also, what do you mean by "unitary convergence"? Do you mean "uniform convergence"?
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Yes, "uniform convergence".
    We don't need x<0. x is always grater than 0 (or x=0)

    Thank you!
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  4. #4
    Senior Member roninpro's Avatar
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    Oops, sorry about that. I didn't see the domain restriction.

    For uniform convergence, did you try looking at the definition? Given \varepsilon>0, is there an N large enough so that n>N implies \left|x\arctan (nx)-\frac{\pi}{2}x\right|<\varepsilon for all x\geq 0?
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  5. #5
    Member mohammadfawaz's Avatar
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    Note that \frac{d}{dx}(x.arctan(n x)) = \frac{nx}{n^2 x^2+1}+arctan(n x).
    Look for the value of x for which the derivative of |x.arctan(nx)-\frac{\pi}{2}x| (for x positive) is zero and draw the table of variations. Now, find the maximum value of that quantity over th given domain. You have uniform convergence if and only if the limit of the value you get is zero.

    hope this helps
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