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Prove that $-x\sin x<\frac{x^4}{6}-x^2$ for all $x\in(0,\pi)$.

My approach is that for similar type of problem I used two function $f(x)=y=-x\sin x$ & $g(x)=y=\frac{x^4}{6}-x^2$

$f(x)-g(x)<0$ for $x\in(0,\pi)$ After this step I am confused.

José Carlos Santos
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Samar Imam Zaidi
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  • I don't see the issue here. With the $f$ and $g$ as proposed by you, you want to prove that $f - g < 0$ on $(0,\pi)$. So of course what you should notice is that $f - g$ is differentiable, and therefore at its point of maximum the derivative is zero. Therefore set the derivative of $f-g$ to $0$, find the points of maxima, and therefore show that the maximum value of $f-g$ is smaller than zero. – Sarvesh Ravichandran Iyer May 26 '18 at 08:13
  • You can simplify the calculations by first dividing the inequality by the (always positive) factor $x$ – Hagen von Eitzen May 26 '18 at 08:14

2 Answers2

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Recall that by Taylor's series we have

$$\sin x >x-\frac16 x^3 \implies -x\sin x<\frac16 x^4-x^2$$

for the proof refer to Proving that $x - \frac{x^3}{3!} < \sin x < x$ for all $x>0$

Following your way we need to show that

$$f(x)=\sin x - x +\frac16 x^2>0$$

then it suffices to note that $f(0)=0$ and that $f(x)$ is strictly ingreasing that is

$$f'(x)=\cos x -1+\frac12 x^2>0$$

for which it suffices to note that $f'(0)=0$ and that $f'(x)$ is strictly ingreasing that is

$$f''(x)=-\sin x+x>0\implies \sin x < x$$

which is true (we can prove also that by continuing the derivation process).

user
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Recall Maclaurin's series of $\sin(x)$, that is

$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}+...+\frac{(-1)^n}{(2n+1)!}x^{2n+1}$$

Then

$$-x\sin x=-x^2+\frac{x^4}{6}-\frac{x^6}{120}+...+\frac{(-1)^{n+1}}{(2n+1)!}x^{2n+2}$$

We rewrite above Maclaurin expansion of $-x \sin x$ as

$$-x\sin x=-x^2+\frac{x^4}{6}-\frac{x^6}{120}+O\left(\frac{x^6}{120}\right)$$

Therefore, for every $x\in(0, \pi)$ inequality holds

$$-x^2+\frac{x^4}{6}-\frac{x^6}{120}+O\left(\frac{x^6}{120}\right)< -x^2+\frac{x^4}{6}$$

This completes the proof.

Petro Kolosov
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