Approximate f(x)=tan x by its degree-3 Maclaurin

Approximate $f(x)=\tan x$ by its degree-3 Maclaurin polynomial and estimate the error for $|x|\lt \tfrac1{10}$.

Step 1: Derivatives of $\tan x$ at $x=0$

Using $f'(x)=1+\tan^2x$, and differentiating again: \[f(0)=0,\quad f'(0)=1,\quad f''(x)=2\tan x(1+\tan^2x)\Rightarrow f''(0)=0,\quad f'''(0)=2. \tag{1}\]

Step 2: Degree-3 Maclaurin polynomial

\[ \tan x = x+\frac{2}{3!}x^3+R_3(x) = x+\frac{x^3}{3}+R_3(x). \tag{2} \]

Step 3: Lagrange remainder and a sharper bound

Directly, $R_3(x)=\dfrac{f^{(4)}(\xi)}{4!}x^4$ with $|\xi|\lt \tfrac1{10}$; a crude bound on $f^{(4)}$ there gives $|R_3(x)|\lt 6.7\times10^{-5}$. A sharper estimate uses that the next nonzero Maclaurin term of $\tan x$ is $\tfrac{2x^5}{15}$, and $|\tan x|\lt \dfrac{|x|}{1-x^2}$ for $|x|\lt \tfrac\pi2$:

\[ |R_3(x)| \le \frac2{15}|x|^5 \lt \frac2{15}\cdot\frac1{10^5} \approx 1.4\times10^{-6}. \tag{3} \]

Step 4: Verification

\[x=0.1\text{: }P(0.1)=0.1+0.000333\overline3=0.10033\overline3,\ \tan(0.1)\approx0.100335,\ |{\rm error}|\approx1.3\times10^{-6}\ \checkmark\]