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Thread: pseudo-sine function?

  1. #1
    Oct 2010

    pseudo-sine function?

    My professor has this exercise in his notes. I may be able to prove it but I don't know what he pseudo-sine function is.
    He states:
    Let $\displaystyle \sigma$ be the pseudo-sine function. Let $\displaystyle f(x)=x\sigma (1/x)$ for $\displaystyle x\not= 0$. Prove $\displaystyle f$ has a removable discontinuity at $\displaystyle 0$.
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  2. #2
    Member HappyJoe's Avatar
    Sep 2010
    The pseudo-sine function might be $\displaystyle \sigma(x) = \sin(\frac{1}{x})$, for $\displaystyle x\neq 0$.

    In this case, we have $\displaystyle f(x) = x\sin(x)$ for $\displaystyle x\neq 0$, and it is not hard to check that this function has a removable discontinuity at $\displaystyle x=0$.

    Do you know how to prove this?

    Also, you better ask your professor, if this is the right definition of the pseudo-sine function (I recall having seen that name somewhere, definitely involving the function $\displaystyle \sin(\frac{1}{x})$, but it might have been $\displaystyle x\sin(\frac{1}{x})$ or something similar).
    Last edited by HappyJoe; Oct 25th 2010 at 09:06 AM. Reason: Beautifying the LaTeX
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  3. #3
    MHF Contributor

    Apr 2005
    A more general concept of "pseudo-sine" is any function that is (quas-) periodic and bounded. Of course, for this proof, you only need "bounded". If $\displaystyle a\le \sigma(x)\le b$ then $\displaystyle ax\le x\sigma(x)\le bx$. Then, as x goes to 0, $\displaystyle 0\le \lim_{x\to 0} x \sigma(x)\le 0$ so the limit exists and is 0. You only need "exists" to prove there is a removable discontinuity at x= 0.
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