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

MATH100 December 2016

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Question 13 (a)
Let $T_{3}(x)$ be the third degree Taylor polynomial centred at $x=0$ for ${\begin{aligned}f(x)&=(x+1)\sin(x).\end{aligned}}$ Write down $T_{3}(x)$. Make sure that you simplify the coefficients.

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Hint
Recall that $T_{3},$ the third-order Taylor polynomial about $0$ for a function $f,$ is given by $T_{3}(x)=f(0)+f'(0)x+{\frac {f''(0)}{2!}}x^{2}+{\frac {f'''(0)}{3!}}x^{3}$.

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. To get $T_{3}(x)$, we first find $f(0),f'(0),f''(0),f''''(0)$. Using product rule, we obtain ${\begin{aligned}f(x)&=(x+1)\sin x&\implies f(0)&=0\\f'(x)&=(x+1)\cos x+\sin x&\implies f'(0)&=1\\f''(x)&=-(x+1)\sin x+\cos x+\cos x=-(x+1)\sin x+2\cos x&\implies f''(0)&=2\\f'''(x)&=-(x+1)\cos x-3\sin x&\implies f'''(0)&=-1\end{aligned}}$ Therefore, $T_{3}(x)=x+{\frac {2}{2!}}x^{2}+{\frac {-1}{3!}}x^{3}=\color {blue}x+x^{2}-{\frac {1}{6}}x^{3}$.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $\color {blue}T_{3}(x)=x+x^{2}-{\frac {x^{3}}{6}}.$