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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.

Archer
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1 Answers1

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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 $x^{3} >6x>6x-6$.

Archer
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Kavi Rama Murthy
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  • Thanks for the answer. I am still not confident about how you proved the inequality – Archer Jul 03 '18 at 06:09
  • @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 $\sin x \geq -1$ – Kavi Rama Murthy Jul 03 '18 at 06:24
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    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. – Theo Bendit Jul 03 '18 at 06:32
  • @TheoBendit Good point! Thanks for pointing it out. – Kavi Rama Murthy Jul 03 '18 at 06:35
  • @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 $n$? – ramanujan Jul 03 '18 at 07:13
  • @ramanujan Please read the comment by Theo Bendit. Absolute convergence is not at all required for my argument. – Kavi Rama Murthy Jul 03 '18 at 07:35