find the taylor polynomials p2(x) and p3(x) centered at a = π for f(x) = sin(x). a. the polynomial p2(x)…

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 $x = a$.

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

The first - derivative of $y = \sin(x)$ is $y'=\cos(x)$, the second - derivative is $y''=-\sin(x)$, the third - derivative is $y'''=-\cos(x)$.

Step3: Evaluate derivatives at $a = \pi$

For $n = 0$, $f(\pi)=\sin(\pi)=0$. For $n = 1$, $f'(\pi)=\cos(\pi)=-1$. For $n = 2$, $f''(\pi)=-\sin(\pi)=0$. For $n = 3$, $f'''(\pi)=-\cos(\pi)=1$.

Step4: Find coefficients for $p_3(x)$

For the Taylor polynomial $p_3(x)=D + E(x-\pi)+F(x - \pi)^2+G(x - \pi)^3$, we have: $D=\frac{f(\pi)}{0!}=0$ (since $f(\pi) = 0$ and $0!=1$). $E=\frac{f'(\pi)}{1!}=\frac{-1}{1}=-1$ (since $f'(\pi)=-1$ and $1!=1$). $F=\frac{f''(\pi)}{2!}=\frac{0}{2}=0$ (since $f''(\pi)=0$ and $2!=2$). $G=\frac{f'''(\pi)}{3!}=\frac{1}{6}$ (since $f'''(\pi)=1$ and $3!=6$).

Answer:

$D = 0$ $E=-1$ $F = 0$ $G=\frac{1}{6}$