Find the Taylor Polynomial $P_{n}( x)$ for $f( x) =e^{x}$ at $0$.

Solution The derivatives of $f( x) =e^{x}$ are $$\begin{array}{rcl@{\qquad}rcl@{\qquad}rcl@{\qquad}rcl@{ }c@{ }rcl} f( x) = e^{x} & f^\prime ( x) = e^{x} & f^{\prime \prime} ( x) = e^{x} & f^{\prime \prime \prime} ( x) = e^{x} & \ldots & f^{( n) }( x) = e^{x} \\ f( 0) = 1 & f^\prime ( 0) = 1 & f^{\prime \prime} ( 0) = 1 & f^{\prime \prime \prime} ( 0) = 1 & \ldots & f^{( n)}( 0) = 1 \end{array}$$

The Taylor Polynomial $P_{n}( x)$ at $0$ is $$\begin{eqnarray*} P_{n}( x) &=&f( 0) +f^\prime ( 0) x+\dfrac{ f^{\prime \prime} ( 0) }{2!}x^{2}+\dfrac{f^{\prime \prime\prime} ( 0) }{3!}x^{3}+\cdots +\dfrac{f^{( n) }( 0) }{n!}x^{n}\\ &=&1+x+\dfrac{x^{2}}{2!}+\dfrac{x^{3}}{3!}+\cdots +\dfrac{ x^{n}}{n!} \end{eqnarray*}$$