# Taylor's polynomial

Now we are going to build a polynomial of degree $n$ that only goes through one given point $x_0$, and its $n$ derivatives coincide in this point with the $n$ derivatives of the original function $f(x)$. This polynomial is named Taylor's polynomial of degree $n$ around the point $x_0$ of the function $f (x)$.

This polynomial is:

$$\begin{array}{c} T_n(x)=f(x_0)+\dfrac{f'(x_0)}{1!}(x-x_0)+ \dfrac{f''(x_0)}{2!}(x-x_0)^2+ \\ \dfrac{f'''(x_0)}{3!}(x-x_0)^3+ \dots \dots +\dfrac{f^{(n)}(x_0)}{n!}(x-x_0)^n \end{array}$$

Therefore, given a $\mathcal{C}^{n+1}$ class function, this means derivable $n +1$ times with continuity, and a point $x_0$ where we know $f(x_0), \ f'(x_0), \ f''(x_0), \dots, \ f^{(n)}(x_0)$, we will be able to calculate Taylor's polynomial.

Let's calculate Taylor's polynomial of order $6$ of the function $f(x)=e^x$ around $x = 0$. We need to know the value of the function and its six first derivatives in the point $x = 0$. Let's calculate them:

$$\begin{array}{lcl} f(0)=1 \\ f'(x)=e^x & \Rightarrow & f'(0)=1 \\ f''(x)=e^x & \Rightarrow & f''(0)=1 \\ f'''(x)=e^x & \Rightarrow & f'''(0)=1 \\ f^{iv}(x)=e^x & \Rightarrow & f^{iv}(0)=1 \\ f^{v}(x)=e^x & \Rightarrow & f^{v}(0)=1 \\ f^{vi}(x)=e^x & \Rightarrow & f^{vi}(0)=1 \end{array}$$

So Taylor's polynomial is:

$$T_6(x)=1+x+\dfrac{1}{2}x^2+\dfrac{1}{6}x^3+\dfrac{1}{24}x^4+ \dfrac{1}{120}x^5+ \dfrac{1}{720}x^6$$

This polynomial can be understood as a polynomial approach of the function $f(x)$. Sometimes it is easier to work with a polynomial than with a complicated function and the Taylor's polynomial becomes very useful. In front of this, we wonder what is our error when using this simplification. Well, this error, which measures how correct is our polynomial compared to the given function, is given by the following expression:

$$\text{error}=\big|f(x)-T_n(x)\big| = \dfrac{f^{(n+1)}(\xi(x))}{(n+1)!}(x-x_0)^{n+1}$$

where $\xi(x)$ is a point between $x_0$ and $x$.