attempt 1: 10 attempts remaining. (a)approximate e^0.5 using the second - order taylor polynomial of…

attempt 1: 10 attempts remaining. (a)approximate e^0.5 using the second - order taylor polynomial of f(x)=e^x centered at a = 0. (do not round until the final answer. then round to five decimal places as needed.) e^0.5≈p2(0.5)= (b) use a calculator to compute the absolute error |e^x - p2(x)| at x = 0.5. (do not round until the final answer. then round to five decimal places as needed.) |e^0.5 - p2(0.5)|=

attempt 1: 10 attempts remaining. (a)approximate e^0.5 using the second - order taylor polynomial of f(x)=e^x centered at a = 0. (do not round until the final answer. then round to five decimal places as needed.) e^0.5≈p2(0.5)= (b) use a calculator to compute the absolute error |e^x - p2(x)| at x = 0.5. (do not round until the final answer. then round to five decimal places as needed.) |e^0.5 - p2(0.5)|=

Answer

Explanation:

Step1: Recall Taylor - polynomial formula

The $n$-th order Taylor polynomial of a function $f(x)$ centered at $a$ is given by $p_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^k$. For $f(x)=e^x$, $f^{(k)}(x)=e^x$ for all $k\geq0$, and $a = 0$. The second - order Taylor polynomial $p_2(x)$ is $p_2(x)=\frac{f(0)}{0!}+\frac{f^{(1)}(0)}{1!}x+\frac{f^{(2)}(0)}{2!}x^2$. Since $f(0)=e^0 = 1$, $f^{(1)}(0)=e^0 = 1$, and $f^{(2)}(0)=e^0 = 1$, then $p_2(x)=1 + x+\frac{x^2}{2}$.

Step2: Approximate $e^{0.5}$

Substitute $x = 0.5$ into $p_2(x)$. We have $p_2(0.5)=1+0.5+\frac{(0.5)^2}{2}=1 + 0.5+\frac{0.25}{2}=1 + 0.5+0.125=1.62500$.

Step3: Calculate the actual value of $e^{0.5}$

Using a calculator, $e^{0.5}\approx1.64872$.

Step4: Compute the absolute error

The absolute error is $|e^{0.5}-p_2(0.5)|=|1.64872 - 1.62500|=0.02372$.

Answer:

(a) $1.62500$ (b) $0.02372$