10.8.3 Find the Taylor polynomial

[karush]

$\textrm{10.8.{7} Find the Taylor polynomial of orders $0, 1, 2$, and $3$ generated by $f$ at $a$.}$

\begin{align*} \displaystyle
f(x)&=\sin{x}
\end{align*}

\[ \begin{array}{llll}\displaystyle
f^0(x)&=\sin{x}&\therefore f^0(\frac{5x}{6})&=\frac{1}{2}\\
\\
f^1(x)&=\cos{x}&\therefore f^1(\frac{5x}{6})&=-\frac{\sqrt{3}}{2}\\
\\
f^2(x)&=-\sin{x}&\therefore f^2(\frac{5x}{6})&=-\frac{1}{2}\\
\\
f^3(x)&=-\cos{x}&\therefore f^3(\frac{5x}{6})&=\frac{\sqrt{3}}{2}\\
\end{array} \]\\

$\textrm{substitute in values}\\$
$\displaystyle
f(x)\approx\frac{\frac{1}{2}}{0!}(x-(\frac{5 \pi}{6}))^{0}
+\frac{- \frac{\sqrt{3}}{2}}{1!}(x-(\frac{5 \pi}{6}))^{1}
+\frac{- \frac{1}{2}}{2!}(x-(\frac{5 \pi}{6}))^{2}
+\frac{\frac{\sqrt{3}}{2}}{3!}(x-(\frac{5 \pi}{6}))^{3}$
$\textrm{substitute in values}$
$\displaystyle
\sin{\left (x \right )}\approx \frac{1}{2}
\frac{\sqrt{3}}{2}(x- \frac{5 \pi}{6})
- \frac{1}{4}(x- \frac{5 \pi}{6})^{2}
+\frac{\sqrt{3}}{12}(x- \frac{5 \pi}{6})^{3}
$
suggestions

hard to know what layTEX to use

 

[Prove It]

$\textrm{10.8.{7} Find the Taylor polynomial of orders $0, 1, 2$, and $3$ generated by $f$ at $a$.}$

\begin{align*} \displaystyle
f(x)&=\sin{x}
\end{align*}

\[ \begin{array}{llll}\displaystyle
f^0(x)&=\sin{x}&\therefore f^0(\frac{5x}{6})&=\frac{1}{2}\\
\\
f^1(x)&=\cos{x}&\therefore f^1(\frac{5x}{6})&=-\frac{\sqrt{3}}{2}\\
\\
f^2(x)&=-\sin{x}&\therefore f^2(\frac{5x}{6})&=-\frac{1}{2}\\
\\
f^3(x)&=-\cos{x}&\therefore f^3(\frac{5x}{6})&=\frac{\sqrt{3}}{2}\\
\end{array} \]\\

$\textrm{substitute in values}\\$
$\displaystyle
f(x)\approx\frac{\frac{1}{2}}{0!}(x-(\frac{5 \pi}{6}))^{0}
+\frac{- \frac{\sqrt{3}}{2}}{1!}(x-(\frac{5 \pi}{6}))^{1}
+\frac{- \frac{1}{2}}{2!}(x-(\frac{5 \pi}{6}))^{2}
+\frac{\frac{\sqrt{3}}{2}}{3!}(x-(\frac{5 \pi}{6}))^{3}$
$\textrm{substitute in values}$
$\displaystyle
\sin{\left (x \right )}\approx \frac{1}{2}
\frac{\sqrt{3}}{2}(x- \frac{5 \pi}{6})
- \frac{1}{4}(x- \frac{5 \pi}{6})^{2}
+\frac{\sqrt{3}}{12}(x- \frac{5 \pi}{6})^{3}
$
suggestions

hard to know what layTEX to use

It's correct, good job :)

 

[HallsofIvy]

I expect that answer is correct but your work is very confusing! You state the question as "find the Taylor series" at x= a. But then you have "[tex]sin\left(\frac{5x}{6}\right)[/tex]". Did you mean [tex]\frac{5\pi}{6}[/tex]? If so where were you told that [tex]a= \frac{5\pi}{6}[/tex].

 

[karush]

I was confused too but that's the way MML gave it.

its a very tedious process easy to make errors