Approximating Functions with 3rd Order Taylor Polynomials

[karush]

$\tiny{242.13.1}$
$\textsf{a. Find the $3^{rd}$ Taylor polynomial for $\sec{x}$ at $a=0$}\\$
\begin{align}
\displaystyle
f^0(x)&=f(x)=\sec{x}\therefore f^0(0)=1\\
&=\frac{1}{0!} x^0=1 \\
f^1(x)&=(\sec{x})'=\tan{x}\sec{x} \therefore f^1(x)=0 \\
&=\frac{1}{0!} x^0+\frac{0}{1!} x^1=1+0=1 \\
f^2(x)&=(\tan{x}\sec{x})'=\left(2\tan^2(x)+1)\sec(x)\right) \therefore f^2(x)=1 \\
&=\frac{1}{0!} x^0
+\frac{0}{1!} x^1
+\frac{1}{2!} x^2=1+0+\frac{1}{2}x^2 \\
&=1+\frac{1}{2}x^2
\end{align}
$\textsf{b. Find the $3^{rd}$
Taylor polynomial for $x^{1/3}$ at $a=8$}\\$
\begin{align}
\displaystyle
f^0(x)&=f(x)= x^{1/3}\therefore f^0(8)=2\\
f^1(x)&=(x^{1/3})'=\frac{1}{3x^{2/3}}
\therefore f^1(8)=\frac{1}{12} \\
f^2(x)&=\left(\frac{1}{3x^{2/3}}\right)''
=\frac{-2}{9x^{5/3}}
\therefore f^2(8)=\frac{-1}{144} \\
f^3(x)&=\left(\frac{-2}{9x^{5/3}}\right)''
=\frac{10}{27x^{8/3}}
\therefore f^3(8)=\frac{5}{3456} \\

f(x)&\approx\frac{2}{0!}(x-8)^{0}
+\frac{\frac{1}{12}}{1!}\left(x-\left(8\right)\right)^{1}
+\frac{- \frac{1}{144}}{2!}\left(x-\left(8\right)\right)^{2}
+\frac{\frac{5}{3456}}{3!}\left(x-\left(8\right)\right)^{3}\\
\sqrt[3]{x}&\approx 2+\frac{1}{12}\left(x-8\right)- \frac{1}{288}\left(x-8\right)^{2}+\frac{5}{20736}\left(x-8\right)^{3}
\end{align}
wasn't sure what was meant by the $3^{rd}$ ?
took time to do this so think is some error

 

[Opalg]

wasn't sure what was meant by the $3^{rd}$ ?
took time to do this so think is some error

Usually when someone writes "$n$th Taylor polynomial" they mean the polynomial going up to and including the term of degree $n$. In general, this polynomial will have $n+1$ terms. So careful authors usually prefer to call it the "$n$th degree Taylor polynomial".

Your calculations both look correct to me, though in a. you should probably have calculated $f^3(x)$ in order to check that $f^3(0) = 0.$

 

[karush]

$\textsf{ok so $\displaystyle 3^{rd}$ would mean
$\displaystyle f^0 + f^1+f^2+f^3$ to be input.}$

 

[karush]

$\tiny{242.13.1}$
$\textsf{a. Find the $3^{rd}$ Taylor polynomial for $\sec{x}$ at $a=0$}\\$
\begin{align}
\displaystyle
f^0(x)&=\sec{x}\therefore f^0(0)=1\\
f^1(x)&=(\sec{x})'=\tan{x}\sec{x} \therefore f^1(0)=0 \\
f^2(x)&=(\tan{x}\sec{x})'=\left(2\tan^2(x)+1)\sec(x)\right) \therefore f^2(0)=1 \\
f^3(x)&=\sec\left(x\right)\tan^3\left(x\right)
+5\sec^3\left(x\right)\tan\left(x\right)=f^3(0)=0 \\
&=\frac{1}{0!} x^0
+\frac{0}{1!} x^1
+\frac{1}{2!} x^2
+\frac{0}{3!} x^3 \\
&=1+0+\frac{1}{2}x^2 +0 \\
sec(x)&=1+\frac{1}{2}x^2
\end{align}
$\textsf{b. Find the $3^{rd}$
Taylor polynomial for $x^{1/3}$ at $a=8$}\\$
\begin{align}
\displaystyle
f^0(x)&=f(x)= x^{1/3}\therefore f^0(8)=2\\
f^1(x)&=(x^{1/3})'=\frac{1}{3x^{2/3}}
\therefore f^1(8)=\frac{1}{12} \\
f^2(x)&=\left(\frac{1}{3x^{2/3}}\right)''
=\frac{-2}{9x^{5/3}}
\therefore f^2(8)=\frac{-1}{144} \\
f^3(x)&=\left(\frac{-2}{9x^{5/3}}\right)''
=\frac{10}{27x^{8/3}}
\therefore f^3(8)=\frac{5}{3456} \\

f(x)&\approx\frac{2}{0!}(x-8)^{0}
+\frac{\frac{1}{12}}{1!}\left(x-\left(8\right)\right)^{1}
+\frac{- \frac{1}{144}}{2!}\left(x-\left(8\right)\right)^{2}
+\frac{\frac{5}{3456}}{3!}\left(x-\left(8\right)\right)^{3}\\
\sqrt[3]{x}&\approx 2+\frac{1}{12}\left(x-8\right)- \frac{1}{288}\left(x-8\right)^{2}+\frac{5}{20736}\left(x-8\right)^{3}
\end{align}