11. 1.33-T nth order Taylor polynomials - - centered at a=100, n=0

[karush]

$\tiny {11. 1.33-T} $
$\textsf{Find the nth order Taylor polynomials of the given function centered at a=100, for $n=0, 1, 2.$}\\$
$$\displaystyle f(x)=\sqrt{x}$$
$\textsf{using}\\$
$$P_n\left(x\right)
\approx\sum\limits_{k=0}^{n}
\frac{f^{(k)}\left(a\right)}{k!}(x-a)^k$$
$\textsf{n=0}\\$
\begin{align}
f^0(x)&= \displaystyle f(x)=\sqrt{x}
\therefore f^0(100)=10\\
P_0\left(x\right)&=\frac{f^0(100)}{0!}(x-100)^{0}= 10
\end{align}
$\textit{just seeing if this is correct }$

 

[Prove It]

$\tiny {11. 1.33-T} $
$\textsf{Find the nth order Taylor polynomials of the given function centered at a=100, for $n=0, 1, 2.$}\\$
$$\displaystyle f(x)=\sqrt{x}$$
$\textsf{using}\\$
$$P_n\left(x\right)
\approx\sum\limits_{k=0}^{n}
\frac{f^{(k)}\left(a\right)}{k!}(x-a)^k$$
$\textsf{n=0}\\$
\begin{align}
f^0(x)&= \displaystyle f(x)=\sqrt{x}
\therefore f^0(100)=10\\
P_0\left(x\right)&=\frac{f^0(100)}{0!}(x-100)^{0}= 10
\end{align}
$\textit{just seeing if this is correct }$

It's correct for P0, now you need P1 and P2...

 

[karush]

$\tiny {11. 1.33-T} $
$\textsf{Find the nth order Taylor polynomials of the given function centered at a=100, for $n=0, 1, 2.$}\\$
$$\displaystyle f(x)=\sqrt{x}$$
$\textsf{using}\\$
$$P_n\left(x\right)
\approx\sum\limits_{k=0}^{n}
\frac{f^{(k)}\left(a\right)}{k!}(x-a)^k$$
$\textsf{n=0}\\$
\begin{align}
f^0(x)&= \displaystyle f(x)=\sqrt{x}
\therefore f^0(100)=10\\
P_0\left(x\right)&=\frac{f^0(100)}{0!}(x-100)^{0}= 10
\end{align}
$\textsf{n=1}\\$
\begin{align}
f^1(x)&= \displaystyle f(x)=\frac{1}{2\sqrt{x}}
\therefore f^1(100)=\frac{1}{20}\\
P_0\left(x\right)&=\frac{f^1(100)}{1!}(x-100)^{1}
= \frac{1}{20}(x-100)
\end{align}
$\textsf{n=2}\\$
\begin{align}
f^2(x)&= \displaystyle f(x)=-\frac{1}{4x^{3/2}}
\therefore f^2(100)=\frac{1}{4000}\\
P_0\left(x\right)&=\frac{f^2(100)}{2!}(x-100)^{2}
= \frac{1}{8000}(x-100)^2
\end{align}
$\textsf{so finally}\\$
$$\sqrt{x}
\approx 10 + \frac{1}{20}(x-100)-\frac{1}{8000}(x-100)^2$$

$\textit{suggestions}$

 

[HallsofIvy]

I think you are misusing the notation "[math]P_n(x)[/math]". Yes, $$P_0(x)= 10$$ because f(100)= 10. And the next term in the Taylor's expansion is [math]\frac{1}{20}[/math] so that $$P_1(x)= \frac{1}{20}(x- 100)+ 10$$. Then the next term is [math]-\frac{1}{8000}(x- 100)^2[/math] so [math]$$P_2(x)= -\frac{1}{8000}(x- 100)^2+ \frac{1}{20}(x- 100)+ 10[/math].$$

 

[karush]

$\tiny {11. 1.33-T} $
$\textsf{Find the nth order Taylor polynomials of the given function centered at a=100, for $n=0, 1, 2.$}\\$
$$\displaystyle f(x)=\sqrt{x}$$
$\textsf{using}\\$
$$P_n\left(x\right)
\approx\sum\limits_{k=0}^{n}
\frac{f^{(k)}\left(a\right)}{k!}(x-a)^k$$
$\textsf{n=0}\\$
\begin{align}
f^0(x)&= \displaystyle f(x)=\sqrt{x}
\therefore f^0(100)=10\\
P_0\left(x\right)&=\frac{f^0(100)}{0!}(x-100)^{0}= 10
\end{align}
$\textsf{n=1}\\$
\begin{align}
f^1(x)&= \displaystyle f(x)=\frac{1}{2\sqrt{x}}
\therefore f^1(100)=\frac{1}{20}\\
P_1\left(x\right)&=\frac{f^1(100)}{1!}(x-100)^{1}
= \frac{1}{20}(x-100)+10
\end{align}
$\textsf{n=2}\\$
\begin{align}
f^2(x)&= \displaystyle f(x)=-\frac{1}{4x^{3/2}}
\therefore f^2(100)=\frac{1}{4000}\\
P_2\left(x\right)&=\frac{f^2(100)}{2!}(x-100)^{2}
= \frac{1}{8000}(x-100)^2+\frac{1}{20}(x-100)+10
\end{align}

$\textit{so would this be 'example' ready?}$