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This is a problem from my calculus / analysis course.

Show that $\forall b > 1$, $\exists x \neq 0$ such that $x = b \sin x$.

My first thought is to use the intermediate value theorem:
Pick any $b>1$.
Let $f(x) = x - b \sin x$. Clearly $f$ is continuous on $\mathbb{R}$.
Then $f(b^2) = b^2 - b \sin (b^2) > 0$.
Now, I need to find some $c > 0$ such that $f(c) < 0$.
Then by the intermediate value theorem, I can show that there exists some $x \in [c,b^2]$ such that $f(x) = 0$, which is the desired result.
However, I cannot find such $c$. Is there any hint on this? Or are there any other approaches to this question?

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    $\begingroup$ Notice, that your function is odd, so $f(-x) = -f(x)$. What happens at $f(-b^2)$? $\endgroup$ Commented May 30, 2022 at 10:58
  • $\begingroup$ $f(0)=0$ and $f'<0$ on a small right neighbourhood of $0$ $\endgroup$ Commented May 30, 2022 at 11:04
  • $\begingroup$ I suppose one can use intermediate value theorem for $f(x)=\frac{\sin x}{x}$ in $x\in(0;\pi)$. $\endgroup$ Commented May 30, 2022 at 12:50

3 Answers 3

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$f(0)=0$ and $f'(0)=1-b<0$. By the definition of the derivative, there exists $x>0$ sufficiently small such that $$ \frac{1-b}2<\frac{f(x)-f(0)}x -(1-b)<\frac{-(1-b)}2 $$ The right inequality can be rewritten $$ \frac{f(x)}x< \frac{1-b}2$$ i.e. $f(x) < \frac{(1-b)x}2 < 0$. Now you can finish with IVT.

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    $\begingroup$ Thank you for your solution! $\endgroup$
    – eurekamath
    Commented May 30, 2022 at 12:38
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Well, since $\lim_{x \rightarrow 0} \frac{sin(x)}{x} = 1$ for every $\epsilon > 0$ you can find an $x_{0}$ such that $sin(x_{0})>x_{0} - \epsilon $. Now use the fact that $b$ is strictly greater than zero. I think this should be enough for you to prove the statement.

If you need more details I can edit the answer, good luck!

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  • $\begingroup$ Honestly I was just trying to give the building blocks of the answer in the most basic way, since he already had the of using the IVT. $\endgroup$
    – Sendobren
    Commented May 30, 2022 at 15:16
  • $\begingroup$ Agreed, sorry for terse comment! $\endgroup$ Commented May 31, 2022 at 0:03
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Using Intermediate Value Theorem: Trivial Solution

We can try solving the question using the Intermediate Value Theorem (here, I obtain the trivial solution $x=0$). Define: $$ f(x) = x-b\sin{x} $$ Note that $f(x)$ is continuous for $x \in \mathbb{R}$. $$ \begin{align} f(b^2) &= b^2 -b\sin{b^2}>0\\ f(-b^2) &= -b^2 -b\sin{-b^2}\\ &= -b^2 +b\sin{b^2} < 0 \end{align} $$ Since $f(x)$ is negative at $-b^2$, postive at $b^2$, and continuous in $[-b^2,b^2]$, by Intermediate Value Theorem: $$ \exists x \in [-b^2,b^2] \text{ such that } f(x) = 0\\ \Rightarrow \exists x \in [-b^2,b^2] \text{ such that }x=b\sin{x} $$ However, as correctly pointed out in the comments, since this $x$ can be zero, it doesn't answer the question exactly. To find a value, not zero, we have to use The Intermediate Value Theorem on two positive values (or two negative values) of $x$.

Using Intermediate Value Theorem: General Solution

We can calculate the value of a $f(x)$ at a value $x=\epsilon>0$ in the $\lim \epsilon \to 0$. $$ \begin{align} f(\epsilon) &= \epsilon - b\sin{\epsilon}\\ \Rightarrow \lim_{\epsilon \to 0} f(\epsilon) &= \epsilon - b\epsilon\\ &=\epsilon(1-b) < 0\text{ since }\epsilon>0\text{ and } b >1 \end{align} $$ We used the identity $\lim_{\epsilon \to 0}\sin{\epsilon} = \epsilon$.

Thus, the value of $f(x)$ is always negative for values of $x$ close to (but greater than zero). We can now use the Intermediate value theorem to show that $x=b\sin{x}$ for some $x>0$.

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    $\begingroup$ but that $x$ can be zero. Need to show a different one. $\endgroup$ Commented May 30, 2022 at 11:14
  • $\begingroup$ this does not answer the question because OP specifically asks for $x\neq 0$. $\endgroup$ Commented May 30, 2022 at 11:15
  • $\begingroup$ You are right, I completely forgot about that. I also didn't use the inequality $b>1$ anywhere. $\endgroup$
    – ananta
    Commented May 30, 2022 at 11:18
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    $\begingroup$ @ananta you used that inequality when you said $f(b^2) > 0$ $\endgroup$ Commented May 30, 2022 at 11:23
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    $\begingroup$ $\lim_{\epsilon\to 0} \sin \epsilon$ cannot be $\epsilon$. $\endgroup$ Commented May 30, 2022 at 12:08

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