(a) find the taylor polynomials up to degree 5 for f(x) = sin(x) centered at a = 0. t0(x) = 0 exactly! t1(x)…

Explanation:

Step1: Recall Taylor - series formula

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

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

• $f(x)=\sin(x)$, $f(0)=\sin(0) = 0$.
• $f^{\prime}(x)=\cos(x)$, $f^{\prime}(0)=\cos(0)=1$.
• $f^{\prime\prime}(x)=-\sin(x)$, $f^{\prime\prime}(0)=-\sin(0) = 0$.
• $f^{(3)}(x)=-\cos(x)$, $f^{(3)}(0)=-\cos(0)=-1$.
• $f^{(4)}(x)=\sin(x)$, $f^{(4)}(0)=\sin(0) = 0$.
• $f^{(5)}(x)=\cos(x)$, $f^{(5)}(0)=\cos(0)=1$.

Step3: Calculate Taylor polynomials

• For $n = 0$: $T_0(x)=\frac{f(0)}{0!}x^0=0$.
• For $n = 1$: $T_1(x)=\frac{f(0)}{0!}x^0+\frac{f^{\prime}(0)}{1!}x^1=0 + 1\times x=x$.
• For $n = 2$: $T_2(x)=\frac{f(0)}{0!}x^0+\frac{f^{\prime}(0)}{1!}x^1+\frac{f^{\prime\prime}(0)}{2!}x^2=0 + x+0=x$.
• For $n = 3$: $T_3(x)=\frac{f(0)}{0!}x^0+\frac{f^{\prime}(0)}{1!}x^1+\frac{f^{\prime\prime}(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3=x+0-\frac{1}{6}x^3=x-\frac{1}{6}x^3$.
• For $n = 4$: $T_4(x)=\frac{f(0)}{0!}x^0+\frac{f^{\prime}(0)}{1!}x^1+\frac{f^{\prime\prime}(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+\frac{f^{(4)}(0)}{4!}x^4=x-\frac{1}{6}x^3+0=x-\frac{1}{6}x^3$.
• For $n = 5$: $T_5(x)=\frac{f(0)}{0!}x^0+\frac{f^{\prime}(0)}{1!}x^1+\frac{f^{\prime\prime}(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+\frac{f^{(4)}(0)}{4!}x^4+\frac{f^{(5)}(0)}{5!}x^5=x-\frac{1}{6}x^3+\frac{1}{120}x^5$.

Answer:

$T_0(x)=0$, $T_1(x)=x$, $T_2(x)=x$, $T_3(x)=x - \frac{1}{6}x^3$, $T_4(x)=x-\frac{1}{6}x^3$, $T_5(x)=x-\frac{1}{6}x^3+\frac{1}{120}x^5$