# Problems from Taylor's polynomial

Find Taylor's polynomial of degree $3$ in $x_0=0$ of the function $f(x)= \dfrac{1}{1+x}$.

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### 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:

$$\left\{ \begin{array}{l} f(x)=\dfrac{1}{1+x}=(1+x)^{-1} \\ f'(x)=-1(1+x)^{-2} \\ f''(x)=2(1+x)^{-3} \\ f'''(x)=-6(1+x)^{-4} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} f(0)=1 \\ f'(0)=-1 \\ f''(0)=2 \\ f'''(0)=-6 \end{array} \right.$$

Therefore, Taylor's polynomial is: $$\begin{array}{rl} T_3(x)=&f(x_0)+\dfrac{f'(x_0)}{1!}(x-x_0)+ \dfrac{f''(x_0)}{2!}(x-x_0)^2+ \dfrac{f'''(x_0)}{3!}(x-x_0)^3 \\ =& 1+\dfrac{-1}{1}(x-0)+\dfrac{2}{2}(x-0)^2+\dfrac{-6}{6}(x-0)^3 \\ =& 1-x+x^2-x^3 \end{array}$$

### Solution:

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

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