Problems from Taylor's polynomial

Development:

To find Taylor's polynomial we have to know the value of the first, second and third derivative of $f(x)$ in the point $x_0=0$.

Let's calculate them:

$$\{\begin{array}{l}f(x)={\displaystyle \frac{1}{1+x}}=(1+x{)}^{-1}\\ f^{\prime }(x)=-1(1+x{)}^{-2}\\ f^{″}(x)=2(1+x{)}^{-3}\\ f^{‴}(x)=-6(1+x{)}^{-4}\end{array}\Rightarrow \{\begin{array}{l}f(0)=1\\ f^{\prime }(0)=-1\\ f^{″}(0)=2\\ f^{‴}(0)=-6\end{array}$$

Therefore, Taylor's polynomial is: $$\begin{array}{rl}{T}_3(x)=& f(x_0)+{\displaystyle \frac{f^{\prime }(x_0)}{1!}}(x-x_0)+{\displaystyle \frac{f^{″}(x_0)}{2!}}(x-x_0{)}^2+{\displaystyle \frac{f^{‴}(x_0)}{3!}}(x-x_0{)}^3\\ =& 1+{\displaystyle \frac{-1}{1}}(x-0)+{\displaystyle \frac{2}{2}}(x-0{)}^2+{\displaystyle \frac{-6}{6}}(x-0{)}^3\\ =& 1-x+x^2-x^3\end{array}$$

Solution:

$T_3(x)=1-x+x^2-x^3$

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