9/23 points details my notes scalcet9 11.11.001. (a) find the taylor polynomials up to degree 5 for f(x) =…
9/23 points details my notes scalcet9 11.11.001. (a) find the taylor polynomials up to degree 5 for f(x) = sin(x) centered at a = 0. t0(x) = 0 exactly! t1(x) = x you got it! t2(x) = x perfect! t3(x) = x - 1/6x^3 nice! t4(x) = x - 1/6x^3 well done. t5(x) = x - 1/6x^3 + 1/120x^5 good work! graph f and these polynomials on a common screen.
Answer
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 Taylor polynomials
- $T_0(x)=\frac{f(0)}{0!}=0$.
- $T_1(x)=\frac{f(0)}{0!}+\frac{f^{\prime}(0)}{1!}x=0 + 1\times x=x$.
- $T_2(x)=\frac{f(0)}{0!}+\frac{f^{\prime}(0)}{1!}x+\frac{f^{\prime\prime}(0)}{2!}x^{2}=x$.
- $T_3(x)=\frac{f(0)}{0!}+\frac{f^{\prime}(0)}{1!}x+\frac{f^{\prime\prime}(0)}{2!}x^{2}+\frac{f^{(3)}(0)}{3!}x^{3}=x-\frac{1}{6}x^{3}$.
- $T_4(x)=\frac{f(0)}{0!}+\frac{f^{\prime}(0)}{1!}x+\frac{f^{\prime\prime}(0)}{2!}x^{2}+\frac{f^{(3)}(0)}{3!}x^{3}+\frac{f^{(4)}(0)}{4!}x^{4}=x-\frac{1}{6}x^{3}$.
- $T_5(x)=\frac{f(0)}{0!}+\frac{f^{\prime}(0)}{1!}x+\frac{f^{\prime\prime}(0)}{2!}x^{2}+\frac{f^{(3)}(0)}{3!}x^{3}+\frac{f^{(4)}(0)}{4!}x^{4}+\frac{f^{(5)}(0)}{5!}x^{5}=x-\frac{1}{6}x^{3}+\frac{1}{120}x^{5}$.
Answer:
$T_0(x)=0$, $T_1(x)=x$, $T_2(x)=x$, $T_3(x)=x-\frac{1}{6}x^{3}$, $T_4(x)=x-\frac{1}{6}x^{3}$, $T_5(x)=x-\frac{1}{6}x^{3}+\frac{1}{120}x^{5}$