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How can I prove that $0\leq{f(x)}<\frac{1}{2}$ for every $x\neq{0}$, when

$$f(x)= \begin{cases} \frac{1-\cos(x)}{x^2}, & x\neq{0} \\ \frac{1}{2}, & x=0 \end{cases} $$

I know how to see that $0\leq{f(x)}$ but I dont know how to prove the other boundary.

Bernard
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codingnight
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    If you don't have the Maclaurin series of $\cos$ available: The function is bounded above by $\frac{2}{x^2}$ which is good enough away from zero. Then it's sufficient, e.g., to show that the function is continuous at $0$. – Travis Willse Mar 04 '18 at 17:12

4 Answers4

2

Note that for $x\neq 0$

$$1-\frac{x^2}2< \cos x \le 1$$

thus

$$0\le \frac{1-\cos x}{x^2}< \frac12$$

Observe that $\cos x>1-\frac{x^2}2$ can be easily shown by MVT.

Using mean value theorem to show that $\cos (x)>1-x^2/2$

user
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$\cos(x) = 1 - x^2/2 + o_{x \to 0}(x^2)$

Hence $f(x) \to_{x \to 0} f(0) = 1/2$

Then you could derivate $f$ on $(0, 1/2)$ to see it is strictly decreasing.

0

Since $1-\cos x = 2 \sin ^2 (x/2)$, replacing $x=2t$, we only need to show $$\left|\frac{\sin t}{t}\right|<1\tag{1}$$ for all $t$. This is sort of classical. Here I only sketch an elementary proof.

Apparently, we only need to show (1) for $|t|<1$, which conveniently lies in $|t|<\pi/2$. Since $\sin t> 0$ for $0<t<\pi/2$ and $\sin(-t) = - \sin t$, we only need to show $$\sin t < t\tag{2}$$ for $0<t<\pi/2$.

Now (2) is easy, given the derivative of $g(t) = \sin t - t$ is $g'(t) = \cos t - 1\le 0$.

Quang Hoang
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Recall that by construction $$0\le\frac{\sin x}x\le1$$ For first-quadrant angles $x$. Then it follows that for first or second quadrant angles, $$0\le\left(\frac{\sin\frac x2}{\frac x2}\right)^2\cos^2\frac x2\le\cos^2\frac x2$$ Equivalently, $$0\le\frac{\sin^2x}{x^2}=\frac{(1+\cos x)(1-\cos x)}{x^2}\le\frac12(1+\cos x)$$ We can verify directly if $x=\frac{\pi}2$ or by division for other first or second quadrant angles that $$0\le f(x)=\frac{1-\cos x}{x^2}\le\frac12$$ Because this is an even function of $x$ it's true for $x\in[-\pi,0)\cup(0,\pi]$ and if we define $f(0)=\frac12$, then it's true for $x=0$ as well. Also it's true for $|x|>\pi$ because then $x^2>4$ and $1-\cos x\le2$. And equality at the upper limit only happens if $\frac{\sin x}x=1$ which is only true in the limit as $x\rightarrow0$ or when $x=(2n+1)\pi$ but we can see by inspection that equality doesn't hold in that case either, so the upper limit is strict except when $x=0$ by the definition of $f(x)$.

user5713492
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