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

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?

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

$$f(0)=0$$ 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.

• Thank you for your solution! Commented May 30, 2022 at 12:38

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!

• 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. Commented May 30, 2022 at 15:16
• Agreed, sorry for terse comment! Commented May 31, 2022 at 0:03

## 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 $$x>0$$.

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