Using Taylor's Theorem to show $|x-tan(x)|\leq 1/300$ for $0\leq x \leq 1/10$

Question

Using Taylor's Theorem deduce that for $0\leq x \leq 1/10$
$|x-tan(x)|\leq 1/300$

So my attempt; to get the taylors theorem about $x_0=0$
$f(x)=x-tan(x)$
$f'(x)=1-sec^{2}(x)$
$f''(x)=-2tan(x)sec^{2}(x)$
$f'''(x)=-2(sec^{2}(x))(1+3tan^{2}(x))$
at $x_0=0$
$f(0)=0$
$f'(0)=0$
$f''(0)=0$
$f'''(0)=-2$
$$f(x)=f(x_0)+f'(x_0)(x-x_0)+(f''(x)(x-x_0)^{2})/2!+(f'''(x)(x-x_0)^{3})/3!+\textrm{remainder term}$$
$$f(x)=0+0+0+(-2(x)^{3}/24)+ \textrm{remainder}$$
for all $0\leq x \leq 1/10$.

Is this correct??$$|(-2(x)^{3}/24)|\leq 1/300$$

Answer 1

Define $f(x) = \tan(x)$ and use the Taylor expansion at $x_0=0$, then we have that $$\tan(x) = x+\frac{x^3}{3}+\frac{2x^5}{15}+r(x)$$ where $r(x)$ is the remainder, satisfying $r(x) = O(x^6)$. Then $|x-\tan(x)| = |\frac{x^3}{3}+\frac{2x^5}{15}+r(x)|$. The big error occurs when $x=\frac{1}{10}$, in that case we have $\left|\frac{1}{3\cdot10^3}+\frac{2}{15\cdot10^5}+O\left(\frac{1}{10^6}\right)\right| < 3.47\cdot10^{-4}+\frac{1}{10^5} = 3.57\cdot10^{-4} < \frac{1}{300}$.

Answer 2

We need an estimate for the function $$f(x):=\tan x-x\qquad(0\leq x\leq {1\over10})\ .$$ As $\tan$ is convex for $0<x<{\pi\over2}$ its Taylor series (with coefficients "out of control") cannot serve for watertight estimates.

By the mean value theorem we have $f(x)-f(0)=x f'(\xi)$ for a $\xi\in\ ]0,x[\ $, and this implies $$0\leq f(x)=x\>\tan^2\xi\leq x\>\tan^2 x\qquad\left(0\leq |x|<{\pi\over2}\right)\ .$$ Now for $0\leq x<{\pi\over2}$ we have $$\sin x\leq x,\quad \cos x\geq1-{x^2\over2}\ ,$$ and this implies $$\tan {1\over10}\leq{{1\over10}\over 1-{1\over200}}={20\over199}\ .$$ It follows that $$0\leq f(x)\leq {1\over10}{400\over 199^2}<{40\over100\cdot 360}={1\over900}\qquad\left(0\leq |x|\leq{1\over10}\right)\ .$$