11. -/23 points details my notes scalcet9 11.11.001. (a) find the taylor polynomials up to degree 5 for f(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), and $f(x)=\sin(x)$.

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

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

Step3: Calculate $T_0(x)$

For $n = 0$, $T_0(x)=\frac{f(0)}{0!}(x - 0)^0=0$.

Step4: Calculate $T_1(x)$

For $n = 1$, $T_1(x)=\frac{f(0)}{0!}(x - 0)^0+\frac{f^{\prime}(0)}{1!}(x - 0)^1=0 + 1\times x=x$.

Step5: Calculate $T_2(x)$

For $n = 2$, $T_2(x)=\frac{f(0)}{0!}(x - 0)^0+\frac{f^{\prime}(0)}{1!}(x - 0)^1+\frac{f^{\prime\prime}(0)}{2!}(x - 0)^2=0 + x+0=x$.

Step6: Calculate $T_3(x)$

For $n = 3$, $T_3(x)=\frac{f(0)}{0!}(x - 0)^0+\frac{f^{\prime}(0)}{1!}(x - 0)^1+\frac{f^{\prime\prime}(0)}{2!}(x - 0)^2+\frac{f^{(3)}(0)}{3!}(x - 0)^3=0 + x+0-\frac{1}{6}x^3=x-\frac{1}{6}x^3$.

Step7: Calculate $T_4(x)$

For $n = 4$, $T_4(x)=\frac{f(0)}{0!}(x - 0)^0+\frac{f^{\prime}(0)}{1!}(x - 0)^1+\frac{f^{\prime\prime}(0)}{2!}(x - 0)^2+\frac{f^{(3)}(0)}{3!}(x - 0)^3+\frac{f^{(4)}(0)}{4!}(x - 0)^4=x-\frac{1}{6}x^3+0=x-\frac{1}{6}x^3$.

Step8: Calculate $T_5(x)$

For $n = 5$, $T_5(x)=\frac{f(0)}{0!}(x - 0)^0+\frac{f^{\prime}(0)}{1!}(x - 0)^1+\frac{f^{\prime\prime}(0)}{2!}(x - 0)^2+\frac{f^{(3)}(0)}{3!}(x - 0)^3+\frac{f^{(4)}(0)}{4!}(x - 0)^4+\frac{f^{(5)}(0)}{5!}(x - 0)^5=x-\frac{1}{6}x^3+\frac{1}{120}x^5$.

Answer:

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