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I have to prove that $f(x)=x\sin (x)$ is surjective in $\Bbb R$.

I thought that I will use the mean value theorem, however I'm finding it hard to do it with sine function. I thought of splitting the function into two indexes, so I could use below $0$ and above $0$ values.

Any ideas?

Thanks,

Alan

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  • $\begingroup$ It is interesting to notice that $f$, as a function from $\mathbb{C}$ to $\mathbb{C}$, is also surjective. $\endgroup$ – Jack D'Aurizio Sep 12 '15 at 15:21
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By surjective I understand $R \rightarrow R$.

We construct sequences $a_n = \frac{\pi}{2}+2\pi n$ and $b_n = -\frac{\pi}{2}-2\pi n$.

$$\lim_{n \rightarrow \infty}f(a_n)=\infty$$

$$\lim_{n \rightarrow \infty}f(b_n)=-\infty$$

EDIT: Surjectivity follows from the Intermediate value theorem using the previous limits and the fact, that $f(x)$ is continuous.

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  • $\begingroup$ Are you sure that you want to use mean value theorem ?? If yes, I would be interested to see how. $\endgroup$ – idm Sep 12 '15 at 14:51
  • $\begingroup$ @idm do you have a better idea? I'd love to hear since I'm having a hard time usually proving that way. $\endgroup$ – Alan Sep 12 '15 at 14:54
  • $\begingroup$ Sorry, i am not native speaker. I thought that was the english name to what i wanted to say :D. $\endgroup$ – MrYouMath Sep 12 '15 at 14:55
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    $\begingroup$ Just for your reference, its called the Intermediate Value Theorem :) $\endgroup$ – Calvin Khor Sep 12 '15 at 14:56
  • $\begingroup$ The fact that $f$ is defined on an interval is crucial here, but you never mention it $\endgroup$ – user261263 Sep 12 '15 at 16:18
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Hint: What is $f(x)$ if $x=(2k+\frac12)\pi$? What if $x=(2k+\frac32)\pi$?

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Let $n \in \mathbb{N}$ and consider $$f\left(\frac{\pi}{2} + 2 \pi n \right) = \frac{\pi}{2} + 2 \pi n$$ This is an increasing subsequence contained in the image of the function.

Let $y \geq 0$. By the Archimedean Principle $$ \exists n \in \mathbb{N} : \frac{y}{2\pi} - \frac{1}{4}\leq n + 1$$ This is equivalent to the statement that $$ \exists n \in \mathbb{N} : y \leq \frac{\pi}{2} + 2 \pi (n+1)$$ Note that $f(0) = 0$. Now let $y \in \mathbb{R} : y \geq 0$ we have $$ 0 \leq y \leq \frac{\pi}{2} + 2 \pi (n + 1) \iff f(0) \leq y \leq f\left(\frac{\pi}{2} + 2 \pi (n+1) \right)$$ By the intermediate value property $$ \exists x \in [0, \frac{\pi}{2} + 2 \pi (n+ 1)] : f(x) = y$$ So we have concluded that the function is surjective for $y \geq 0$.

Finally in order to deal with the $y < 0$ case we use a similar proof but instead consider the subsequence $$ f\left(-\frac{\pi}{2} + 2 \pi n \right) = -\frac{\pi}{2} + 2 \pi n$$

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