attempt 1: 10 attempts remaining. find the taylor polynomials p2(x) and p4(x) centered at a = π for…

attempt 1: 10 attempts remaining. find the taylor polynomials p2(x) and p4(x) centered at a = π for f(x)=sin(x). a. the polynomial p2(x) must be in the form a + b(x - π/1)+c(x - π/1)^2 where a = b = and c = b. the polynomial p4(x) must be in the form

attempt 1: 10 attempts remaining. find the taylor polynomials p2(x) and p4(x) centered at a = π for f(x)=sin(x). a. the polynomial p2(x) must be in the form a + b(x - π/1)+c(x - π/1)^2 where a = b = and c = b. the polynomial p4(x) must be in the form

Answer

Explanation:

Step1: Recall Taylor - series formula

The Taylor series of a function $f(x)$ centered at $a$ is given by $f(x)=\sum_{n = 0}^{\infty}\frac{f^{(n)}(a)}{n!}(x - a)^n$, where $f^{(n)}(a)$ is the $n$-th derivative of $f(x)$ evaluated at $x = a$. For a second - degree Taylor polynomial $p_2(x)=A + B(x - a)+C(x - a)^2$, we have $A=f(a)$, $B = f^{\prime}(a)$, and $C=\frac{f^{\prime\prime}(a)}{2!}$. Here $a=\pi$ and $f(x)=\sin(x)$.

Step2: Calculate $A$

First, find $f(a)$ where $a = \pi$. Since $f(x)=\sin(x)$, then $f(\pi)=\sin(\pi)=0$. So $A = 0$.

Step3: Calculate the first - derivative

The first - derivative of $f(x)=\sin(x)$ is $f^{\prime}(x)=\cos(x)$. Evaluate it at $x = \pi$, so $f^{\prime}(\pi)=\cos(\pi)=- 1$. Thus $B=-1$.

Step4: Calculate the second - derivative

The second - derivative of $f(x)=\sin(x)$ is $f^{\prime\prime}(x)=-\sin(x)$. Evaluate it at $x = \pi$, so $f^{\prime\prime}(\pi)=-\sin(\pi)=0$. Then $C=\frac{f^{\prime\prime}(\pi)}{2!}=\frac{0}{2}=0$.

Answer:

$A = 0$, $B=-1$, $C = 0$