What is a bound of $\frac{1}{\sin x} - \frac{1}{x}$on $(0,a]$?

Question

Is $\dfrac{1}{\sin x} - \dfrac{1}{x}$ bounded in $(0,a]$, $a<\frac\pi2$?

How can I prove this? Any hints or suggestions?

Answer 1

Lemma: For $|t|\le \frac\pi2$ we have, $$\left|\frac{1}{\sin t}-\frac{1}{t}\right|\le 1-\frac2\pi$$

From this we have that for for all $0\le t\le\frac{\pi}{2}$ $$\tan t \sin t \ge t^2 \Longleftrightarrow -\frac{\cos t}{\sin^2 t}+\frac{1}{t^2} =\left(\frac{1}{\sin t}-\frac{1}{t}\right)'\ge0 $$

Therefore, the function, $f:t\mapsto \frac{1}{\sin t}-\frac{1}{t}$ is increasing on $(0,\frac\pi2)$

then for $t\in(0,\frac\pi2)$ $$f(0)\le f(t)\le f(\frac\pi2)$$ that is, $$ 0=f(0)=\lim_{h \to 0}\left(\frac{1}{\sin h}-\frac{1}{h}\right)\le \left(\frac{1}{\sin t}-\frac{1}{t}\right)\le f(\frac\pi2)=1-\frac2\pi $$

The result holds since the function is odd it suffices to consider $-t\in(0,\frac\pi2).$

Answer 2

Consider $$ \frac{1}{\sin x}-\frac{1}{x}= \frac{x-\sin x}{x^3}\frac{x}{\sin x}x $$ This implies that $$ \lim_{x\to0}\left(\frac{1}{\sin x}-\frac{1}{x}\right)=\dotsb $$ and, since the function is continuous over $(0,a]$, so long as $0<a<\pi$,…

Note that $a$ cannot be $\ge\pi$ or the function would not be defined on the interval $(0,a]$.

Answer 3

Assuming that $a < \pi$, the answer is yes. In this case, the function is clearly bounded on $[\varepsilon, a]$ for any $\epsilon > 0$ since it is continuous there and the set is compact. The only question is then whether it remains bounded in a neighborhood of zero; or equivalently, does $$\lim_{x\to 0^+} \left(\frac{1}{\sin(x)} - \frac 1 x \right)$$ exist as a finite number. We show that this limit does exist. Indeed, \begin{align*}\lim_{x\to 0^+} \left(\frac{1}{\sin(x)} - \frac 1 x \right) = \lim_{x \to 0^+} \frac{\sin(x) - x}{x\sin(x)}\end{align*} and using a Taylor expansion, the numerator is $O(x^3)$ whereas the denominator is $O(x^2)$ as $x \to 0^+$ so the limit is zero.

Answer 4

Ready for an overkill? $\frac{1}{\sin z}$ is a meromorphic function with simple poles at $\pi\mathbb{Z}$. Since $\operatorname*{Res}_{z=0}\frac{1}{z}=1$, the function $g(z)=\frac{1}{\sin z}-\frac{1}{z}$ (re-defined at the origin as zero) is holomorphic in the region $|z|<\pi$, hence it is obviously bounded over the interval $\left[0,\frac{\pi}{2}\right]$. Additionally $$ \frac{1}{\sin z}-\frac{1}{z}=\sum_{n\geq 1}\frac{2z(-1)^n}{z^2-\pi^2 n^2}=\sum_{n\geq 1}\frac{2\,\zeta(2n)\left(1-\frac{2}{4^n}\right)}{\pi^{2n}}\,z^{2n-1} $$ gives us all the coefficients of the MacLaurin series of the LHS, proving it is a positive, increasing and convex function on $(0,\pi)$.

Answer 5

For $0\leq z$ we have $$0\leq \sin z\leq z.$$ Integrating with respect to $z$ from $z=0$ to $z=y\geq 0$ we have $$0\leq 1-\cos y\leq y^2/2 \quad \text { for } y\geq 0.$$ Integrating with respect to $y$ from $y=0$ to $y=x\geq 0$ we have $$0\leq x+\sin x\leq x^3/6 \quad \text { for } x\geq 0.$$ Therefore $x\geq 0\implies x-x^3/6\leq \sin x \leq x.$

For $0<x<\pi /2$ let $\sin x =x-x^3\delta /6$ where $0\leq \delta \leq 1.$ Then $$\frac {1}{x}-\frac {1}{\sin x}=\frac {-x^3\delta /6}{x^2(1-x^3\delta /6}= \frac {-x\delta /6} {1-x^2\delta /6}.$$ The numerator in the last expression above belongs to the interval $(-\pi /12,0)$ and the denominator is at least $1-(\pi /2)^2/6=M>0.$ So for $0<x\leq\pi /2$ we have $$\frac {-\pi /12}{M}\leq \frac {1}{x}-\frac {1}{\sin x}\leq 0.$$ Not the sharpest estimate but sufficient for the Q.

Remark. Continuing the sequence of integrations in the first part of this A we obtain $x\geq 0\implies S(x,2n)\leq \sin x \leq S(x,2n-1)$ where $S(x,m)=\sum_{j=0}^m x^{2j+1}(-1)^j/(2j+1)!$ From that, and from $\sin (-x)=-\sin x,$ we obtain the power series for $\sin x$ for all $x\in \Bbb R.$.............. I learned this from the book "$101$ Great Problems In Elementary Mathematics" by Heinrich Dorrie.

Answer 6

The only worrying point is obviously $x=0^+$.

We know that for $x>0$

$$x>\sin x>x-\frac{x^3}6$$ so that

$$\frac1x<\frac1{\sin x}<\frac1x\frac6{6-x^2}$$ and

$$0<\frac1{\sin x}-\frac1x<\frac1x\left(\frac6{6-x^2}-1\right)=\frac x{6-x^2}$$

and the function is indeed bounded.

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By the way, this shows that the slope at the origin does not exceed $\dfrac16$.