Prove $\sin(x) > x - \frac{x^3}{3!} $ on $(0, \sqrt{20})$

Question

I'm having a bit of trouble with this because my attempted proof breaks down.

Proof: It is sufficient to show that $f(x) = \sin(x) - x + \frac{x^3}{3!} > 0$ on $I = (0, \sqrt{20})$. This is true if $f'(x)$ is strictly increasing on the interval and $f(0) \geq 0$. We can apply this property on the first and second derivatives as well. We note $f^{(3)}(x) = -\cos(x) + 1$. However I can only show that $f^{(3)}(x) \geq 0$ on $I$ since at $x = \frac{3\pi}{2}$ $f^{(3)}(x) = 0$

Answer 1

You are on the right track. The function $f(x) = \sin(x) - x + \frac{x^3}{3!}$ satisfies $f(0) = f'(0) = f''(0) = 0$ and $$ f^{(3)}(x) = -\cos(x) + 1 \ge 0 $$ with equality only at the points $x_k = \frac \pi 2 + 2 k\pi$, $k \in \Bbb Z$. It follows that $f''$ is strictly increasing on each interval $[x_k, x_{k+1}]$ Therefore $f''$ is strictly increasing on $\Bbb R$ and strictly positive on $(0, \infty)$.

Now you can conclude that $f'$ and consequently $f$ are strictly increasing on $[0,\infty)$, and therefore $f(x) > 0$ for all $x > 0$.

Answer 2

Well, consider the series expansion of $\sin(x)$: $$ \sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots $$ So from your definition of $f(x)$, $$ f(x)=\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots\geq \frac{x^5}{5!}-\frac{x^7}{7!} $$ We can solve the inequality $$ \frac{x^5}{5!}-\frac{x^7}{7!}>0\Leftrightarrow42-x^2>0\Leftrightarrow x<\sqrt{42} $$ since $x>0$.

So when $0<x<\sqrt{42}$, your inequality is true.

I actually don't know where the $\sqrt{20}$ come from in the question. Anyone have a clue?

Answer 3

You can prove it directly: $$f(x)=\sin(x) - x + \frac{x^3}{3!}>0,x>0;\\ f'(x)=\cos (x) -1+\frac{x^2}{2}>0 \iff \left(\frac x2\right)^2>\sin ^2 \left(\frac x2\right) \iff \\ \left(\frac x2-\sin \frac x2\right)\left(\frac x2+\sin \frac x2\right)>0 \iff \\ \frac x2-\sin \frac x2>0 \iff \frac x2>\sin \frac x2,x>0$$ Note:

1) $\cos (x)-1=-2\sin ^2 \left(\frac x2\right)$.

2) $x>\sin x, x>0$ is true, because $g(x)=x-\sin x$ is strictly increasing.

Answer 4

For $x\ge0$, integrating from $0$ at each step: $$ \cos(x)\le1\\ \Downarrow\\ \sin(x)\le x\\ \Downarrow\\ 1-\cos(x)\le\tfrac12x^2\\ \Downarrow\\ x-\sin(x)\le\tfrac16x^3\\ $$ Therefore, for all $x\ge0$, $$ \sin(x)\ge x-\tfrac16x^3 $$