206.11.3.27 first three nonzero terms of the Taylor series

[karush]

$\textsf{a. Find the first three nonzero terms
of the Taylor series $a=\frac{3\pi}{4}$}$
\begin{align}
\displaystyle
f^0(x)&=\sin{x} &\therefore \ \ f^0(a)&=\sin{x} \\
f^1(x)&=\cos{x} &\therefore \ \ f^1(a)&= -\frac{\sqrt{2}}{2}\\
f^2(x)&=- \sin{x}&\therefore \ \ f^2(a)&=\frac{\sqrt{2}}{2} \\
\end{align}

$\textsf{so then}$
\begin{align}
\displaystyle
f\left(x\right)&
\approx\frac{\frac{\sqrt{2}}{2}}{0!}\left(x-\left(\frac{3 \pi}{4}\right)\right)^{0}+\frac{- \frac{\sqrt{2}}{2}}{1!}\left(x-\left(\frac{3 \pi}{4}\right)\right)^{1}+\frac{- \frac{\sqrt{2}}{2}}{2!}\left(x-\left(\frac{3 \pi}{4}\right)\right)^{2}\\
\sin{\left (x \right )}&\approx
\frac{\sqrt{2}}{2}
- \frac{\sqrt{2}}{2}\left(x- \frac{3 \pi}{4}\right)
- \frac{\sqrt{2}}{4}\left(x- \frac{3 \pi}{4}\right)^{2}
\end{align}

 

[Prove It]

Re: 206.11.3.27 first three nonzero terms of the Taylor serie

$\textsf{a. Find the first three nonzero terms
of the Taylor series $a=\frac{3\pi}{4}$}$
\begin{align}
\displaystyle
f^0(x)&=\sin{x} &\therefore \ \ f^0(a)&=\sin{x} \\
f^1(x)&=\cos{x} &\therefore \ \ f^1(a)&= -\frac{\sqrt{2}}{2}\\
f^2(x)&=- \sin{x}&\therefore \ \ f^2(a)&=\frac{\sqrt{2}}{2} \\
\end{align}

$\textsf{so then}$
\begin{align}
\displaystyle
f\left(x\right)&
\approx\frac{\frac{\sqrt{2}}{2}}{0!}\left(x-\left(\frac{3 \pi}{4}\right)\right)^{0}+\frac{- \frac{\sqrt{2}}{2}}{1!}\left(x-\left(\frac{3 \pi}{4}\right)\right)^{1}+\frac{- \frac{\sqrt{2}}{2}}{2!}\left(x-\left(\frac{3 \pi}{4}\right)\right)^{2}\\
\sin{\left (x \right )}&\approx
\frac{\sqrt{2}}{2}
- \frac{\sqrt{2}}{2}\left(x- \frac{3 \pi}{4}\right)
- \frac{\sqrt{2}}{4}\left(x- \frac{3 \pi}{4}\right)^{2}
\end{align}

Well you have calculated $\displaystyle \begin{align*} f^{(0)}(a) \end{align*}$ and $\displaystyle \begin{align*} f^{(2)}(a) \end{align*}$ incorrectly, but somehow written them correctly in the final answer...

 

[karush]

Re: 206.11.3.27 first three nonzero terms of the Taylor serie

$\textsf{a. Find the first three nonzero terms
of the Taylor series $a=\frac{3\pi}{4}$}$
\begin{align}
\displaystyle
f^0(x)&=\sin{x} &\therefore \ \ f^0(a)&=\frac{\sqrt{2}}{2} \\
f^1(x)&=\cos{x} &\therefore \ \ f^1(a)&= -\frac{\sqrt{2}}{2}\\
f^2(x)&=- \sin{x}&\therefore \ \ f^2(a)&=-\frac{\sqrt{2}}{2} \\
\end{align}

$\textsf{so then}$
\begin{align}
\displaystyle
f\left(x\right)&
\approx\frac{\frac{\sqrt{2}}{2}}{0!}\left(x-\left(\frac{3 \pi}{4}\right)\right)^{0}+\frac{- \frac{\sqrt{2}}{2}}{1!}\left(x-\left(\frac{3 \pi}{4}\right)\right)^{1}+\frac{- \frac{\sqrt{2}}{2}}{2!}\left(x-\left(\frac{3 \pi}{4}\right)\right)^{2}\\
\sin{\left (x \right )}&\approx
\frac{\sqrt{2}}{2}
- \frac{\sqrt{2}}{2}\left(x- \frac{3 \pi}{4}\right)
- \frac{\sqrt{2}}{4}\left(x- \frac{3 \pi}{4}\right)^{2}
\end{align}