Finding a Taylor Polynomial for \(f( x) =e^{x}\)

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 \[ \begin{eqnarray*} P_{n}( x) &=&f( 0) +f^\prime ( 0) x+\dfrac{ f^{\prime \prime} ( 0) }{2!}x^{2}+\dfrac{f^{\prime \prime\prime} ( 0) }{3!}x^{3}+\cdots +\dfrac{f^{( n) }( 0) }{n!}x^{n}\\ &=&1+x+\dfrac{x^{2}}{2!}+\dfrac{x^{3}}{3!}+\cdots +\dfrac{ x^{n}}{n!} \end{eqnarray*} \]