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I looked at your answer to the question posted How to prove $\sin(1/x)$ is not uniformly continuous

thank you for your helpful explanation on how to think about it. But I am failing to see why we will choose $x=\frac{1}{2πk−π/2}$ . I know that we want something $\sin(1/x)=1$ and $\sin(1/y)=−1$, but how did you get that answer. Would you mind elucidating on your motivation?

Also, I am struggling to get an intuitive idea to work with continuous uniform problems. I understand the definition, but am failing to work with it.

Thanks

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user43901
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  • With those choices, $\sin(\frac{1}{x})$ oscillates between -1 and 1. – Euler....IS_ALIVE Oct 07 '12 at 16:17
  • This question is also relevant: http://math.stackexchange.com/questions/206371/tricks-to-intuitively-see-uniform-continuity – Michael Hardy Oct 07 '12 at 16:28
  • But when I solved the equation, I did not get the x values quoted in the question. If you would be kind enough to kindly walk me through them once, I think I will understand the concept compared to just getting the answer. – user43901 Oct 07 '12 at 17:37

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I think you are referring to Jonas Meyer's answer which says:

"For what $x$ and $y$ is it true that $\sin\left(\frac{1}{x}\right)=−1$ and $\sin\left(\frac{1}{y}\right)=1$?"

Already this may be a problem as he was working with $x$ such that $\sin\left(\frac{1}{x}\right) = - 1$ but in your question, you have written $\sin\left(\frac{1}{x}\right) = 1$ (and vice versa for $y$). Is this why you don't get the stated values?

In Jonas Meyer's comment he says that $x = \displaystyle\frac{1}{2\pi k - \frac{\pi}{2}}$ and $y = \displaystyle\frac{1}{2\pi k + \frac{\pi}{2}}$. I have included the working for the $x$ values below if you are still unable to obtain them.

$$\displaystyle\sin\left(\frac{1}{x}\right) = -1 \Leftrightarrow \frac{1}{x} = 2\pi k - \frac{\pi}{2},\ k \in \mathbb{Z} \Leftrightarrow x = \frac{1}{2\pi k - \frac{\pi}{2}},\ k \in \mathbb{Z}.$$

You should be able to find the $y$ values in a similar way.

Michael Albanese
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    Thank you so much! This really helps. By the way, how did you include the flash in effect to your answer? It looks really cool! Yes, I figured out the error. – user43901 Oct 25 '12 at 02:51
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    You're welcome. To get the grey box, use > followed by text; to get the grey box and the hidden text, use >! followed by text. – Michael Albanese Oct 25 '12 at 03:25