Question

Show that:
\[\sin x < x - \dfrac{{{x^3}}}{6} + \dfrac{{{x^5}}}{{120}}\] for $x > 0$.

Answer 1

Hint: To prove this inequality, we will use the Rolle's theorem. The Rolle's theorem is as follows: If $f:[a,b] \to \mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$ such that $f(a) = f(b)$, then there exists some $c \in (a,b)$ such that $f'(c) = 0$.

Complete step by step answer:
We will first consider the following function,
\[f(x) = x - \dfrac{{{x^3}}}{6} + \dfrac{{{x^5}}}{{120}} - \sin x\]
To show that
\[\sin x < x - \dfrac{{{x^3}}}{6} + \dfrac{{{x^5}}}{{120}}\]
is the same as having to show that $f(x) > 0$ for $x > 0$. We can see that $f(0) = 0$. Now, we take the first derivative of $f(x)$,
\[f'(x) = 1 - \dfrac{{{x^2}}}{2} + \dfrac{{{x^4}}}{{24}} - \cos x\]
and we can directly see that $f'(0) = 0$.
Now, we will look at the second derivative of $f(x)$,
\[f''(x) = - x + \dfrac{{{x^3}}}{6} + \sin x\]
Let $g(x) = f''(x)$. We can see that at $x = 0$, \[g(x),g'(x),g''(x),g'''(x)\] and ${g^{(4)}}(x)$ are $0$. This means that the first four derivatives of $g(x)$ vanish at $x=0$.
Now, let ${x_1} > 0$ be such that $g({x_1}) = 0$, that means, ${{x}_{1}}$ is a positive value where the function $g$ is $0$. Now we have $g(x)=g({{x}_{1}})$. So now we can apply Rolle's theorem to the function $g$, and to its subsequent derivatives. Calculating the higher derivatives of the function $g$, we get the fourth derivative as ${g^{(4)}}(x) = \sin x$, and by properties of the sine function, we can safely say that our function does not change signs for $0 < x < \pi$.
Now, we can see that $g(x) = f''(x) > 0$ for $x > 0$ because $\sin x \geqslant - 1$ and $- x + \dfrac{{{x^3}}}{6} > \dfrac{3}{2}$.
Hence, $f(x) > 0$ for $x > 0$.

Note:
Understanding when the Rolle's theorem or the mean value theorem can be applied is important. The trick part in this question is to look at the higher derivatives and apply Rolle's theorem. This inequality that we proved for $x>0$ does not hold true if we consider $x<0$.