calculate the taylor polynomials t2(x) and t3(x) centered at x = a for f(x) = 4 sin(x), a = π/2. (express…

Explanation:

Step1: Recall Taylor - polynomial formula

The Taylor polynomial of degree $n$ for a function $f(x)$ centered at $x = a$ is given by $T_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^k$, where $f^{(k)}(x)$ is the $k$-th derivative of $f(x)$.

Step2: Find derivatives of $f(x)=4\sin(x)$

First - derivative: $f^{\prime}(x)=4\cos(x)$. Second - derivative: $f^{\prime\prime}(x)=-4\sin(x)$. Third - derivative: $f^{\prime\prime\prime}(x)=-4\cos(x)$.

Step3: Evaluate derivatives at $a=\frac{\pi}{2}$

$f(\frac{\pi}{2}) = 4\sin(\frac{\pi}{2})=4$. $f^{\prime}(\frac{\pi}{2}) = 4\cos(\frac{\pi}{2}) = 0$. $f^{\prime\prime}(\frac{\pi}{2})=-4\sin(\frac{\pi}{2})=-4$. $f^{\prime\prime\prime}(\frac{\pi}{2})=-4\cos(\frac{\pi}{2}) = 0$.

Step4: Calculate $T_2(x)$

$T_2(x)=\frac{f(\frac{\pi}{2})}{0!}(x-\frac{\pi}{2})^0+\frac{f^{\prime}(\frac{\pi}{2})}{1!}(x - \frac{\pi}{2})^1+\frac{f^{\prime\prime}(\frac{\pi}{2})}{2!}(x-\frac{\pi}{2})^2$. Substituting the values: $T_2(x)=4+0\times(x - \frac{\pi}{2})+\frac{-4}{2}(x-\frac{\pi}{2})^2=4 - 2(x-\frac{\pi}{2})^2$.

Step5: Calculate $T_3(x)$

$T_3(x)=\frac{f(\frac{\pi}{2})}{0!}(x-\frac{\pi}{2})^0+\frac{f^{\prime}(\frac{\pi}{2})}{1!}(x - \frac{\pi}{2})^1+\frac{f^{\prime\prime}(\frac{\pi}{2})}{2!}(x-\frac{\pi}{2})^2+\frac{f^{\prime\prime\prime}(\frac{\pi}{2})}{3!}(x-\frac{\pi}{2})^3$. Substituting the values: $T_3(x)=4+0\times(x - \frac{\pi}{2})+\frac{-4}{2}(x-\frac{\pi}{2})^2+0\times(x-\frac{\pi}{2})^3=4 - 2(x-\frac{\pi}{2})^2$.

Answer:

$T_2(x)=4 - 2(x-\frac{\pi}{2})^2$ $T_3(x)=4 - 2(x-\frac{\pi}{2})^2$