1. We start with creating a table of the derivatives of \(e^x\) evaluated at \(x=0\).

\[

\begin{array}{lll}

f(x)=e^x & \Rightarrow & f(0)=1 \\

f^{\prime}(x)=e^x & \Rightarrow & f^{\prime}(0)=1 \\

f^{\prime \prime}(x)=e^x & \Rightarrow & f^{\prime \prime}(0)=1 \\

\vdots & & \vdots \\

f^{(n)}(x)=e^x & \Rightarrow & f^{(n)}(0)=1

\end{array}

\]

By the definition of the Maclaurin series, we have

\[

\begin{aligned}

p_n(x) &=f(0)+f^{\prime}(0) x+\frac{f^{\prime \prime}(0)}{2 !} x^2+\frac{f^{\prime \prime \prime}(0)}{3 !} x^3+\cdots+\frac{f^n(0)}{n !} x^n \\

&=1+x+\frac{1}{2} x^2+\frac{1}{6} x^3+\frac{1}{24} x^4+\cdots+\frac{1}{n !} x^n .

\end{aligned}

\]

2. Using our answer from part 1, we have

\[

p_5(x)=1+x+\frac{1}{2} x^2+\frac{1}{6} x^3+\frac{1}{24} x^4+\frac{1}{120} x^5 .

\]

To approximate the value of \(e\), note that \(e=e^1=f(1) \approx p_5(1)\). It is very straightforward to evaluate \(p_5(1)\) :

\[

p_5(1)=1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}=\frac{163}{60} \approx 2.71667 .

\]