pseudo-sine function?

**zebra2147** · October 25th 2010, 07:16 AM

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 $\sigma$ be the pseudo-sine function. Let $f(x)=x\sigma (1/x)$ for $x\not= 0$ . Prove $f$ has a removable discontinuity at $0$ .

**HappyJoe** · October 25th 2010, 09:05 AM

The pseudo-sine function might be $\sigma(x) = \sin(\frac{1}{x})$ , for $x\neq 0$ .

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

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 $\sin(\frac{1}{x})$ , but it might have been $x\sin(\frac{1}{x})$ or something similar).

**HallsofIvy** · October 25th 2010, 04:58 PM

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 $a\le \sigma(x)\le b$ then $ax\le x\sigma(x)\le bx$ . Then, as x goes to 0, $0\le \lim_{x\to 0} x \sigma(x)\le 0$ 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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