0 like 0 dislike
25 views
1. Find the $n^{\text {th }}$ Maclaurin polynomial for $f(x)=e^x$.
2. Use $p_5(x)$ to approximate the value of $e$.
| 25 views

0 like 0 dislike
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 $p_5(1)$ :
$p_5(1)=1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}=\frac{163}{60} \approx 2.71667 .$
by Platinum (141,884 points)

0 like 0 dislike
2 like 0 dislike
2 like 0 dislike
1 like 0 dislike
1 like 0 dislike
2 like 0 dislike
2 like 0 dislike
2 like 0 dislike
2 like 0 dislike