find the taylor polynomial t₂(x) and compute the error |f(x) - t₂(x)| for the given values of a and x…

Explanation:

Step1: Recall Taylor - polynomial formula

The second - degree Taylor polynomial $T_2(x)$ of a function $f(x)$ about $x = a$ is given by $T_2(x)=f(a)+f^{\prime}(a)(x - a)+\frac{f^{\prime\prime}(a)}{2!}(x - a)^2$. First, find the derivatives of $y = f(x)=e^{\sin(x)}$. Using the chain - rule, if $y = e^{u}$ and $u=\sin(x)$, then $y^\prime=f^{\prime}(x)=e^{\sin(x)}\cos(x)$. And $y^{\prime\prime}=f^{\prime\prime}(x)=e^{\sin(x)}\cos^{2}(x)-e^{\sin(x)}\sin(x)=e^{\sin(x)}(\cos^{2}(x)-\sin(x))$.

Step2: Evaluate function and its derivatives at $a=\frac{\pi}{2}$

When $a = \frac{\pi}{2}$, $f(\frac{\pi}{2})=e^{\sin(\frac{\pi}{2})}=e^{1}=e\approx2.718282$, $f^{\prime}(\frac{\pi}{2})=e^{\sin(\frac{\pi}{2})}\cos(\frac{\pi}{2})=e\times0 = 0$, $f^{\prime\prime}(\frac{\pi}{2})=e^{\sin(\frac{\pi}{2})}(\cos^{2}(\frac{\pi}{2})-\sin(\frac{\pi}{2}))=e(0 - 1)=-e$.

Step3: Construct the Taylor polynomial

$T_2(x)=f(\frac{\pi}{2})+f^{\prime}(\frac{\pi}{2})(x-\frac{\pi}{2})+\frac{f^{\prime\prime}(\frac{\pi}{2})}{2}(x - \frac{\pi}{2})^2$. Substituting the values, we get $T_2(x)=e+0\times(x - \frac{\pi}{2})-\frac{e}{2}(x - \frac{\pi}{2})^2$. When $x = 1.5$, $T_2(1.5)=e-\frac{e}{2}(1.5-\frac{\pi}{2})^2$. $1.5-\frac{\pi}{2}\approx1.5 - 1.570796=-0.070796$, $(1.5-\frac{\pi}{2})^2\approx0.005012$, $T_2(1.5)=e-\frac{e}{2}\times0.005012\approx2.718282- \frac{2.718282}{2}\times0.005012\approx2.711468$.

Step4: Calculate the error

First, find $f(1.5)=e^{\sin(1.5)}$. $\sin(1.5)\approx0.997495$, $f(1.5)=e^{0.997495}\approx2.710117$. The error $|f(1.5)-T_2(1.5)|=|2.710117 - 2.711468|= 0.001351\approx1.35\times10^{-3}$.

Answer:

$|f(1.5)-T_2(1.5)| = 1.35\times10^{-3}$