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Use ${\left(1+x\right)}^{1\text{/}3}=1+\frac{1}{3}x-\frac{1}{9}{x}^{2}+\frac{5}{81}{x}^{3}-\frac{10}{243}{x}^{4}+\text{\cdots}$ with $x=1$ to approximate ${2}^{1\text{/}3}.$
Use the approximation ${\left(1-x\right)}^{2\text{/}3}=1-\frac{2x}{3}-\frac{{x}^{2}}{9}-\frac{4{x}^{3}}{81}-\frac{7{x}^{4}}{243}-\frac{14{x}^{5}}{729}+\text{\cdots}$ for $\left|x\right|<1$ to approximate ${2}^{1\text{/}3}={2.2}^{\mathrm{-2}\text{/}3}.$
Twice the approximation is $1.260\text{\u2026}$ whereas ${2}^{1\text{/}3}=\mathrm{1.2599.}...$
Find the $25\text{th}$ derivative of $f\left(x\right)={\left(1+{x}^{2}\right)}^{13}$ at $x=0.$
Find the $99$ th derivative of $f\left(x\right)={\left(1+{x}^{4}\right)}^{25}.$
${f}^{\left(99\right)}\left(0\right)=0$
In the following exercises, find the Maclaurin series of each function.
$f\left(x\right)=x{e}^{2x}$
$f\left(x\right)={2}^{x}$
$\sum _{n=0}^{\infty}\frac{{\left(\text{ln}\phantom{\rule{0.1em}{0ex}}\left(2\right)x\right)}^{n}}{n\text{!}}$
$f\left(x\right)=\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{x}$
$f\left(x\right)=\frac{\text{sin}\left(\sqrt{x}\right)}{\sqrt{x}},\phantom{\rule{0.5em}{0ex}}\left(x>0\right),$
For $x>0,\text{sin}\left(\sqrt{x}\right)={\displaystyle \sum _{n=0}^{\infty}{\left(\mathrm{-1}\right)}^{n}\frac{{x}^{\left(2n+1\right)\text{/}2}}{\sqrt{x}\left(2n+1\right)\text{!}}}={\displaystyle \sum _{n=0}^{\infty}{\left(\mathrm{-1}\right)}^{n}\frac{{x}^{n}}{\left(2n+1\right)\text{!}}}.$
$f\left(x\right)=\text{sin}\left({x}^{2}\right)$
$f\left(x\right)={e}^{{x}^{3}}$
${e}^{{x}^{3}}={\displaystyle \sum _{n=0}^{\infty}\frac{{x}^{3n}}{n\text{!}}}$
$f\left(x\right)={\text{cos}}^{2}x$ using the identity ${\text{cos}}^{2}x=\frac{1}{2}+\frac{1}{2}\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}\left(2x\right)$
$f\left(x\right)={\text{sin}}^{2}x$ using the identity ${\text{sin}}^{2}x=\frac{1}{2}-\frac{1}{2}\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}\left(2x\right)$
${\text{sin}}^{2}x=\text{\u2212}{\displaystyle \sum _{k=1}^{\infty}\frac{{\left(\mathrm{-1}\right)}^{k}{2}^{2k-1}{x}^{2k}}{\left(2k\right)\text{!}}}$
In the following exercises, find the Maclaurin series of $F\left(x\right)={\displaystyle {\int}_{0}^{x}f\left(t\right)\phantom{\rule{0.1em}{0ex}}dt}$ by integrating the Maclaurin series of $f$ term by term. If $f$ is not strictly defined at zero, you may substitute the value of the Maclaurin series at zero.
$F\left(x\right)={\displaystyle {\int}_{0}^{x}{e}^{\text{\u2212}{t}^{2}}}dt;f\left(t\right)={e}^{\text{\u2212}{t}^{2}}={\displaystyle \sum _{n=0}^{\infty}{\left(\mathrm{-1}\right)}^{n}}\frac{{t}^{2n}}{n\text{!}}$
$F\left(x\right)={\text{tan}}^{\mathrm{-1}}x;\phantom{\rule{0.5em}{0ex}}f\left(t\right)=\frac{1}{1+{t}^{2}}={\displaystyle \sum _{n=0}^{\infty}{\left(\mathrm{-1}\right)}^{n}{t}^{2n}}$
${\text{tan}}^{\mathrm{-1}}x={\displaystyle \sum _{k=0}^{\infty}\frac{{\left(\mathrm{-1}\right)}^{k}{x}^{2k+1}}{2k+1}}$
$F\left(x\right)={\text{tanh}}^{\mathrm{-1}}x;\phantom{\rule{0.5em}{0ex}}f\left(t\right)=\frac{1}{1-{t}^{2}}={\displaystyle \sum _{n=0}^{\infty}{t}^{2n}}$
$F\left(x\right)={\text{sin}}^{\mathrm{-1}}x;\phantom{\rule{0.5em}{0ex}}f\left(t\right)=\frac{1}{\sqrt{1-{t}^{2}}}={\displaystyle \sum _{k=0}^{\infty}\left(\begin{array}{c}\frac{1}{2}\hfill \\ k\hfill \end{array}\right)\frac{{t}^{2k}}{k\text{!}}}$
${\text{sin}}^{\mathrm{-1}}x={\displaystyle \sum _{n=0}^{\infty}\left(\begin{array}{c}\frac{1}{2}\hfill \\ n\hfill \end{array}\right)\frac{{x}^{2n+1}}{\left(2n+1\right)n\text{!}}}$
$F\left(x\right)={\displaystyle {\int}_{0}^{x}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}t}{t}dt};\phantom{\rule{0.5em}{0ex}}f\left(t\right)=\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}t}{t}={\displaystyle \sum _{n=0}^{\infty}{\left(\mathrm{-1}\right)}^{n}\frac{{t}^{2n}}{\left(2n+1\right)\text{!}}}$
$F\left(x\right)={\displaystyle {\int}_{0}^{x}\text{cos}\left(\sqrt{t}\right)\phantom{\rule{0.1em}{0ex}}dt};\phantom{\rule{0.5em}{0ex}}f\left(t\right)={\displaystyle \sum _{n=0}^{\infty}{\left(\mathrm{-1}\right)}^{n}\frac{{x}^{n}}{\left(2n\right)\text{!}}}$
$F\left(x\right)={\displaystyle \sum _{n=0}^{\infty}{\left(\mathrm{-1}\right)}^{n}\frac{{x}^{n+1}}{\left(n+1\right)\left(2n\right)\text{!}}}$
$F\left(x\right)={\displaystyle {\int}_{0}^{x}\frac{1-\text{cos}\phantom{\rule{0.1em}{0ex}}t}{{t}^{2}}dt};\phantom{\rule{0.5em}{0ex}}f\left(t\right)=\frac{1-\text{cos}\phantom{\rule{0.1em}{0ex}}t}{{t}^{2}}={\displaystyle \sum _{n=0}^{\infty}{\left(\mathrm{-1}\right)}^{n}\frac{{t}^{2n}}{\left(2n+2\right)\text{!}}}$
$F\left(x\right)={\displaystyle {\int}_{0}^{x}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}\left(1+t\right)}{t}dt};\phantom{\rule{0.5em}{0ex}}f\left(t\right)={\displaystyle \sum _{n=0}^{\infty}{\left(\mathrm{-1}\right)}^{n}\frac{{t}^{n}}{n+1}}$
$F\left(x\right)={\displaystyle \sum _{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}\frac{{x}^{n}}{{n}^{2}}}$
In the following exercises, compute at least the first three nonzero terms (not necessarily a quadratic polynomial) of the Maclaurin series of $f.$
$f\left(x\right)=\text{sin}\left(x+\frac{\pi}{4}\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{cos}\left(\frac{\pi}{4}\right)+\text{cos}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{sin}\left(\frac{\pi}{4}\right)$
$f\left(x\right)=\text{tan}\phantom{\rule{0.1em}{0ex}}x$
$x+\frac{{x}^{3}}{3}+\frac{2{x}^{5}}{15}+\text{\cdots}$
$f\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}\left(\text{cos}\phantom{\rule{0.1em}{0ex}}x\right)$
$f\left(x\right)={e}^{x}\text{cos}\phantom{\rule{0.1em}{0ex}}x$
$1+x-\frac{{x}^{3}}{3}-\frac{{x}^{4}}{6}+\text{\cdots}$
$f\left(x\right)={e}^{\text{sin}\phantom{\rule{0.1em}{0ex}}x}$
$f\left(x\right)={\text{sec}}^{2}x$
$1+{x}^{2}+\frac{2{x}^{4}}{3}+\frac{17{x}^{6}}{45}+\text{\cdots}$
$f\left(x\right)=\text{tanh}\phantom{\rule{0.1em}{0ex}}x$
$f\left(x\right)=\frac{\text{tan}\sqrt{x}}{\sqrt{x}}$ (see expansion for $\text{tan}\phantom{\rule{0.1em}{0ex}}x)$
Using the expansion for $\text{tan}\phantom{\rule{0.1em}{0ex}}x$ gives $1+\frac{x}{3}+\frac{2{x}^{2}}{15}.$
In the following exercises, find the radius of convergence of the Maclaurin series of each function.
$\text{ln}\phantom{\rule{0.1em}{0ex}}\left(1+x\right)$
$\frac{1}{1+{x}^{2}}$
$\frac{1}{1+{x}^{2}}={\displaystyle \sum _{n=0}^{\infty}{\left(\mathrm{-1}\right)}^{n}{x}^{2n}}$ so $R=1$ by the ratio test.
${\text{tan}}^{\mathrm{-1}}x$
$\text{ln}\phantom{\rule{0.1em}{0ex}}\left(1+{x}^{2}\right)$
$\text{ln}\phantom{\rule{0.1em}{0ex}}\left(1+{x}^{2}\right)={\displaystyle \sum _{n=1}^{\infty}\frac{{\left(\mathrm{-1}\right)}^{n-1}}{n}}{x}^{2n}$ so $R=1$ by the ratio test.
Find the Maclaurin series of $\text{sinh}\phantom{\rule{0.1em}{0ex}}x=\frac{{e}^{x}-{e}^{\text{\u2212}x}}{2}.$
Find the Maclaurin series of $\text{cosh}\phantom{\rule{0.1em}{0ex}}x=\frac{{e}^{x}+{e}^{\text{\u2212}x}}{2}.$
Add series of ${e}^{x}$ and ${e}^{\text{\u2212}x}$ term by term. Odd terms cancel and $\text{cosh}\phantom{\rule{0.1em}{0ex}}x={\displaystyle \sum _{n=0}^{\infty}\frac{{x}^{2n}}{\left(2n\right)\text{!}}}.$
Differentiate term by term the Maclaurin series of $\text{sinh}\phantom{\rule{0.1em}{0ex}}x$ and compare the result with the Maclaurin series of $\text{cosh}\phantom{\rule{0.1em}{0ex}}x.$
[T] Let ${S}_{n}\left(x\right)={\displaystyle \sum _{k=0}^{n}{\left(\mathrm{-1}\right)}^{k}\frac{{x}^{2k+1}}{\left(2k+1\right)\text{!}}}$ and ${C}_{n}\left(x\right)={\displaystyle \sum _{n=0}^{n}{\left(\mathrm{-1}\right)}^{k}\frac{{x}^{2k}}{\left(2k\right)\text{!}}}$ denote the respective Maclaurin polynomials of degree $2n+1$ of $\text{sin}\phantom{\rule{0.1em}{0ex}}x$ and degree $2n$ of $\text{cos}\phantom{\rule{0.1em}{0ex}}x.$ Plot the errors $\frac{{S}_{n}\left(x\right)}{{C}_{n}\left(x\right)}-\text{tan}\phantom{\rule{0.1em}{0ex}}x$ for $n=1,\mathrm{..},5$ and compare them to $x+\frac{{x}^{3}}{3}+\frac{2{x}^{5}}{15}+\frac{17{x}^{7}}{315}-\text{tan}\phantom{\rule{0.1em}{0ex}}x$ on $\left(-\frac{\pi}{4},\frac{\pi}{4}\right).$
The ratio
$\frac{{S}_{n}\left(x\right)}{{C}_{n}\left(x\right)}$ approximates
$\text{tan}\phantom{\rule{0.1em}{0ex}}x$ better than does
${p}_{7}\left(x\right)=x+\frac{{x}^{3}}{3}+\frac{2{x}^{5}}{15}+\frac{17{x}^{7}}{315}$ for
$N\ge 3.$ The dashed curves are
$\frac{{S}_{n}}{{C}_{n}}-\text{tan}$ for
$n=1,2.$ The dotted curve corresponds to
$n=3,$ and the dash-dotted curve corresponds to
$n=4.$ The solid curve is
${p}_{7}-\text{tan}\phantom{\rule{0.1em}{0ex}}x.$
Use the identity $2\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x=\text{sin}\left(2x\right)$ to find the power series expansion of ${\text{sin}}^{2}x$ at $x=0.$ ( Hint: Integrate the Maclaurin series of $\text{sin}\left(2x\right)$ term by term.)
If $y={\displaystyle \sum _{n=0}^{\infty}{a}_{n}{x}^{n}},$ find the power series expansions of $x{y}^{\prime}$ and ${x}^{2}y\text{\u2033}.$
By the term-by-term differentiation theorem, ${y}^{\prime}={\displaystyle \sum _{n=1}^{\infty}n{a}_{n}{x}^{n-1}}$ so ${y}^{\prime}={\displaystyle \sum _{n=1}^{\infty}n{a}_{n}{x}^{n-1}}\phantom{\rule{0.2em}{0ex}}x{y}^{\prime}={\displaystyle \sum _{n=1}^{\infty}n{a}_{n}{x}^{n}},$ whereas ${y}^{\prime}={\displaystyle \sum _{n=2}^{\infty}n\left(n-1\right){a}_{n}{x}^{n-2}}$ so $xy\text{\u2033}={\displaystyle \sum _{n=2}^{\infty}n\left(n-1\right){a}_{n}{x}^{n}}.$
[T] Suppose that $y={\displaystyle \sum _{k=0}^{\infty}{a}_{k}{x}^{k}}$ satisfies ${y}^{\prime}=\mathrm{-2}xy$ and $y\left(0\right)=0.$ Show that ${a}_{2k+1}=0$ for all $k$ and that ${a}_{2k+2}=\frac{\text{\u2212}{a}_{2k}}{k+1}.$ Plot the partial sum ${S}_{20}$ of $y$ on the interval $[\mathrm{-4},4].$
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