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Show that: $\sin x> x- \dfrac{x^3}{6} ~~\forall ~x>0$

Attempt:

Let $y = \sin x - x+\dfrac{x^3}{6}$

We have to prove that y is increasing for $x >0$

$y' = \cos x -1 + \dfrac{x^2}{2}$

$y'' =-\sin x+x$

$y''>0$ for all $x>0$.

$\implies y'$ is increasing for all $x>0$

$\implies y$ is an increasing function with $y(0+)>0$

Hence proved.

Attempt 2:

Writing the Taylor expansion of $\sin x$, we get this inequality to be proven:

$\dfrac{x^5}{5!}- \dfrac{x^7}{7!}+ \dfrac{x^9}{9!}+...>0$

Is it possible to prove it? I tried to (by rearranging) but couldn't.

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  • $\begingroup$ There are several posts about the question in the title - such as Proving that $x - \frac{x^3}{3!} < \sin x < x$ for all $x>0$ and other posts linked there. However it seems that your main question is the inequality $\frac{x^5}{5!}- \frac{x^7}{7!}+ \frac{x^9}{9!}+...>0$ near the end of your question. If this is what you want to ask, maybe you should put this inequality in the title. $\endgroup$ – Martin Sleziak Jul 3 '18 at 6:03
  • $\begingroup$ @Abcd I hope my answer is acceptable now. Let me know if you have further questions. $\endgroup$ – Kavi Rama Murthy Jul 3 '18 at 6:08
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Consider any alternating series $a_1-a_2+a_3...$ with $a_n >0$ and $a_n$ decreasing to $0$. If the series is absolutely convergent then we can group the terms as $(a_1-a_2)+(a_3-a_4)+...$ and the sum is greater than $a_1-a_2$ because the other terms are non-negative. Your question is a special case of this. Attempt 2 also works by the same method. Details: $x^{5} /5! -x^{7} /7! >0$ if $x <\sqrt {42}$ Similarly $x^{7} /7! -x^{9} /9! >0$, etc for such $x$ so we get $\sin x > x- x^{3} /3!$. It remains to see $\sin x > x-x^{3} /3!$ when $x \geq \sqrt {42}$. But here $x-x^{3} /3!<-1 \leq \sin x$ since $x^{3} >6x>6x-6$.

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  • $\begingroup$ Thanks for the answer. I am still not confident about how you proved the inequality $\endgroup$ – Archer Jul 3 '18 at 6:09
  • $\begingroup$ @Abcd In either attempt you have to consider small values of $x$ and large values separately. My idea is that for small values you can get the required inequalities by grouping the terms 2 by 2. For large values it seems to be necessary to use the fact that $\sin x \geq -1$ $\endgroup$ – Kavi Rama Murthy Jul 3 '18 at 6:24
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    $\begingroup$ Small note: absolute convergence isn't necessary here. The partial sums of the series $(a_1 - a_2) + (a_3 - a_4) + \ldots$ are just a subsequence of the partial sums of $a_1 - a_2 + a_3 - a_4 + \ldots$. In general, conditionally convergent sums are associative, even if they're not commutative. $\endgroup$ – Theo Bendit Jul 3 '18 at 6:32
  • $\begingroup$ @TheoBendit Good point! Thanks for pointing it out. $\endgroup$ – Kavi Rama Murthy Jul 3 '18 at 6:35
  • $\begingroup$ @Kavi Rama Murthy how can we say that ' if the series is absolutely convergent, other terms are non negative?' Is it necessary for convergent alternating series that $a_{n+1} \leq a_n$ for all $n$? $\endgroup$ – ramanujan Jul 3 '18 at 7:13

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