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Prove that $|\sin x-x |\le \frac{1}{6}|x|^3$

I've been trying to solve this problem for the last hour but I just can't crack it. I know I'm supposed to use Cauchy's Mean Value Theorem somehow but can't figure out how.

Edit: This problem is from chapter 4 of Foundations of Analysis by Taylor. The book hasn't covered Taylor series yet. I am supposed solve this using a method that is already covered. It is very likely that the intended solution involves the usage of Cauchy's form of the mean value theorem.

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    $\begingroup$ I guess you haven't yet studied Taylor series, power sieres orstuff? $\endgroup$
    – DonAntonio
    Commented Jul 18, 2019 at 13:01
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    $\begingroup$ I have studied Taylor series last year. However, this problem is from the 4h chapter of Foundations of Analysis. The book hasn't covered Taylor's formula yet. I am supposed to solve this using Cauchy's form of the mean value theorem. $\endgroup$
    – cppcoder
    Commented Jul 18, 2019 at 13:34

3 Answers 3

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A solution using Cauchy's Mean-Value Theorem: With $$ f(x) = \sin(x) - x, g(x) = x^3 $$ we have for all $x > 0$ $$ \frac{\sin(x)-x}{x^3} = \frac{f(x)-f(0)}{g(x)-g(0)} = \frac{f'(c)}{g'(c)} = \frac{\cos(c)-1}{3c^2} $$ for some $c$ between $0$ and $x$. Now apply Cauchy's Mean-Value Theorem again twice to conclude that this is $$ \begin{align} &= \frac{-\sin(d)}{6d} \quad \text{for some $d$ between $0$ and $c$} \\ &= \frac{-\cos(e)}{6} \quad \text{for some $e$ between $0$ and $d$} \end{align} $$ and therefore $$ \left| \frac{\sin(x)-x}{x^3} \right| = \frac{|\cos(e)|}{6} \le \frac 16 \, . $$

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Assume $x \ge 0$.

Set $f(x) = \sin x-x$. We have $f'(x) = \cos x-1 \le 0$ so $f$ is decreasing on $[0,+\infty)$. Therefore $f(x) \le f(0) = 0$ so $\sin x \le x$.

Set $$g(x) = 1-\cos x - \frac12x^2$$ We have $g'(x) = \sin x -x\le 0$ so $g$ is decreasing on $[0,+\infty)$. Therefore $g(x) \le g(0) = 0$ so $1-\cos x \le \frac12x^2$.

Finally, set $$h(x) = x-\sin x - \frac16x^3$$ We have $$h'(x)= 1-\cos x - \frac12x^2 \le 0$$ so $h$ is decreasing on $[0,+\infty)$. Therefore $h(x) \le h(0) = 0$ so $x-\sin x \le \frac16x^3$.

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    $\begingroup$ How do you know that $1 - cos(x) \le x^2/2$? It's true but your link just quotes it as well. It seems assuming that is equivalent to the original problem. $\endgroup$
    – quarague
    Commented Jul 18, 2019 at 13:07
  • $\begingroup$ @quarague Set $g(x) = 1-\cos x - \frac12 x^2$. We have $g'(x) = \sin x - x \le 0$ so $g$ is decreasing and thus $g(x) \le g(0) = 0$. $\endgroup$
    – mechanodroid
    Commented Jul 18, 2019 at 13:08
  • $\begingroup$ So if you start with $\sin(x) \le x$ for $x \ge 0$, you can use your method to prove the original statement. One could prove $\sin(x) \le x$ using power series but then one can prove the initial problem that way. Maybe there is a way without power series to prove that as well. Note the absolute value signs in the initial problem, I don't think this can be assumed as given. $\endgroup$
    – quarague
    Commented Jul 18, 2019 at 13:13
  • $\begingroup$ @quarague Set $h(x) = \sin x - x$. Then $h'(x) = \cos x - 1 \le 0$ so $h$ is decreasing and thus $h(x) \le h(0) = 0$. $\endgroup$
    – mechanodroid
    Commented Jul 18, 2019 at 13:18
  • $\begingroup$ This answer seems to be circular in the required methods to be used... $\endgroup$
    – DonAntonio
    Commented Jul 18, 2019 at 13:21
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When $0\leq x\leq4$ then $$\sin x-x=-{x^3\over6}+{x^5\over120}-{x^7\over5040}+\ldots\ ,$$ and therefore $|\sin x-x|\leq{x^3\over6}$ since the series on the RHS is alternating, and in the given $x$-domain the terms are decreasing in absolute value. When $x\geq4$ we can estimate $$|\sin x-x|\leq 1+x\leq{1\over6}x^3\ ,$$ since the rightmost term is obviously abounding when $x\geq4$.

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