attempt 1: 10 attempts remaining. imagine that you need to compute e^0.4 but you have no calculator or other…

Explanation:

Step1: Recall Taylor - series formula

The Taylor series of $f(x)=e^{x}$ about $x = a$ is given by $\sum_{n = 0}^{\infty}\frac{f^{(n)}(a)}{n!}(x - a)^{n}$. For $a = 0$, $f(x)=e^{x}$, $f^{(n)}(x)=e^{x}$, so $f^{(n)}(0)=e^{0}=1$. The second - degree Taylor polynomial $P_{2}(x)$ of $e^{x}$ about $x = 0$ is $P_{2}(x)=1 + x+\frac{x^{2}}{2}$.

Step2: Recall the error bound formula

The error bound for the Taylor polynomial $P_{n}(x)$ of a function $f(x)$ is given by $|R_{n}(x)|\leq\frac{M}{(n + 1)!}|x - a|^{n+1}$, where $M$ is an upper - bound for $|f^{(n + 1)}(t)|$ on the interval between $a$ and $x$. Here, $n = 2$, $a = 0$, $x=0.4$, and $f(x)=e^{x}$, so $f^{(3)}(t)=e^{t}$. Since $t\in[0,0.4]$ and $e^{t}$ is an increasing function and $e^{0.4}<e < 3$, we can take $M = 3$.

Step3: Calculate the error bound

Substitute $n = 2$, $M = 3$, $a = 0$, and $x = 0.4$ into the error - bound formula $|R_{2}(x)|\leq\frac{M}{(n + 1)!}|x - a|^{n + 1}$. We get $|R_{2}(0.4)|\leq\frac{3}{3!}|0.4-0|^{3}$. Since $3!=6$, then $\frac{3}{3!}|0.4|^{3}=\frac{3}{6}\times(0.4)^{3}=\frac{1}{2}\times0.064 = 0.032$.

Answer:

$0.032$