MHB 206.11.3.39 Find the first four nonzero terms of the Taylor series

[karush]

$\tiny{206.11.3.39}$
$\textsf{a. Find the first four nonzero terms
of the Taylor series $a=0$}$

\begin{align}
\displaystyle
f^0(x)&=(1+x)^{-2} &\therefore \ \ f^0(a)&= 1 \\
f^1(x)&=\frac{-2}{(x+1)^3} &\therefore \ \ f^1(a)&= -2 \\
f^2(x)&=\frac{6}{(x+1)^4} &\therefore \ \ f^2(a)&= 6 \\
f^3(x)&=\frac{-24}{(x+1)^5} &\therefore \ \ f^3(a)&= -24 \\
\end{align}
$\textsf{so then}$
\begin{align}
\displaystyle
f\left(x\right)&\approx\frac{1}{0!}x^{0}
+\frac{-2}{1!}x^{1}+\frac{6}{2!}x^{2}+\frac{-24}{3!}x^{3} \\
f\left(x\right)&\approx 1-2x+3x^{2}-4x^{3}
\end{align}

$\textsf{b. approximate $\frac{1 }{1,14^{-2}}$ }\\$

$\textsf{not sure where this is plugged in??}$

 

[topsquark]

$\textsf{b. approximate $\frac{1 }{1,14^{-2}}$ }\\$

At a guess are you asking about [math](1 + 0.14)^{-2}[/math] ?

In that case you are looking at x = 0.14

-Dan

 

[karush]

ok I see..
was going to plug the whole thing in..

but if $x=1.14$ then

$$f(x)\approx 1-2(1.14)+3(1.14)^2-4(1.14)^3\approx 0.77$$

 

[topsquark]

ok I see..
was going to plug the whole thing in..

but if $x=1.14$ then

$$f(x)\approx 1-2(1.14)+3(1.14)^2-4(1.14)^3\approx 0.77$$

Look at your problem more carefully. You have the Taylor series for [math](1 + x)^{-2}[/math] for x close to 0. You want to approximate [math](1 + 0.14)^{-2}[/math]. Comparing the two expressions you can see that we want x = 0.14, not 1.14.

-Dan

 

[karush]

$\textsf{Ok i put $0.14$ in the TI so $f(x)\approx 0.77$ but didn't in the post.}$

 

[Prove It]

$\displaystyle \begin{align*} f(x) &= \frac{1}{\left( 1 + x \right) ^2} \end{align*}$

Notice that $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\,\left( -\frac{1}{1 + x} \right) = \frac{1}{\left( 1 + x \right) ^2} \end{align*}$ and

$\displaystyle \begin{align*} -\frac{1}{1 + x} &= -\frac{1}{1 - \left( -x \right) } \\ &= -\sum_{n = 0}^{\infty}{ \left( -x \right) ^n } \textrm{ for } \left| -x \right| < 1 \implies \left| x \right| < 1 \\ &= -\sum_{n = 0}^{\infty}{ \left( -1 \right) ^n \, x^n } \\ &= \sum_{n = 0}^{\infty}{ \left( -1 \right) ^{n + 1}\,x^n } \\ \\ \frac{\mathrm{d}}{\mathrm{d}x}\,\left( -\frac{1}{1 + x} \right) &= \frac{\mathrm{d}}{\mathrm{d}x} \,\left[ \sum_{n = 0}^{\infty}{ \left( -1 \right) ^{n + 1}\,x^n } \right] \\ \frac{1}{\left( 1 + x \right)^2} &= \sum_{n = 0}^{\infty}{ \left( -1 \right) ^{n+ 1}\,n\,x^{n - 1} } \\ &= 0 + 1 - 2\,x + 3\,x^2 - 4\,x^3 + \dots \textrm{ for } \left| x \right| < 1 \end{align*}$

so the first four nonzero terms are $\displaystyle \begin{align*} 1 - 2\,x + 3\,x^2 - 4\,x^3 \end{align*}$.