Chapter 9 - Power Series - 9.1 Approximating Functions with Polynomials - 9.1 Exercises - Page 673: 64

$0.1196$

The remainder in the n-th order Taylor's polynomial $T_n(x)$ for a function $f(x)$ centered at $a$ can be expressed as: $|R_n(x)|=|f(x)-T_n(x)| \leq M \dfrac{|x-a|^{n+1}}{(n+1)!}$ We are given that $f(x)=\tan x$ and $T(x)=x$ and $n=2$ Thus, we have $|R_n(x)|\leq 5 \times \dfrac{|\dfrac{\pi}{6}-0|^{2+1}}{(2+1)!} \approx 0.1196$