Science:Math Exam Resources/Courses/MATH100/December 2016/Question 13 (a)/Solution 2

We know (or can calculate) that the Taylor expansion of $\sin(x)$ centred at $x=0$ is

$\sin(x)=x-{\frac {x^{3}}{3!}}+\cdots .$

Hence the Taylor expansion of $f(x)=(x+1)\sin(x)$ centred at $x=0$ is

$(x+1)\sin(x)=(x+1)\left(x-{\frac {x^{3}}{3!}}+\cdots \right)=x^{2}-{\frac {x^{4}}{3!}}+x-{\frac {x^{3}}{3!}}+\cdots .$

Discarding the terms of degree higher than 3, we obtain the third degree Taylor polynomial

$\color {blue}T_{3}(x)=x+x^{2}-{\frac {x^{3}}{6}}.$