Science:Math Exam Resources/Courses/MATH104/December 2016/Question 03 (b)/Solution 1

As you see in the Hint, to find the Taylor polynomial of degree 3 of $f$, we first need to find $f(1)$, $f'(1)$, $f''(1)$, and $f'''(1)$.

Plugging $x=1$ into $f(x)$, we get $f(1)=1\cdot \ln 1=0$.

Then taking derivative, we have $f'(x)=\ln x+x\cdot {\frac {1}{x}}=1+\ln x$, so that $f'(1)=1$.

Taking more derivatives, we obtain

$f''(x)={\frac {1}{x}}\implies f''(1)=1$

and

$f'''(x)=-{\frac {1}{x^{2}}}\implies f'''(1)=-1$

Therefore, the 3rd degree Taylor polynomial is:

${\begin{aligned}T_{3}(x)&=f(1)+{\frac {f'(1)}{1!}}(x-1)+{\frac {f''(1)}{2!}}(x-1)^{2}+{\frac {f^{(3)}(1)}{3!}}(x-1)^{3}\\&=0+{\frac {1}{1!}}(x-1)+{\frac {1}{2!}}(x-1)^{2}+{\frac {-1}{3!}}(x-1)^{3}\\&=x-1+{\frac {1}{2}}(x-1)^{2}-{\frac {1}{6}}(x-1)^{3}\end{aligned}}$

Answer: $\color {blue}T_{3}(x)=x-1+{\frac {1}{2}}(x-1)^{2}-{\frac {1}{6}}(x-1)^{3}$