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


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.

  • $\begingroup$ 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

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

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.