# Find the Maclaurin series for f(x)=x^4cos x. Also, find the third term for this series. The Maclaurin series for cos(x) is given by: cos(x)=sum_{n=0}^infty(-1)^nfrac{x^{2n}}{(2n)!}=1-frac{x^2}{2!}+frac{x^4}{4!}-frac{x^6}{6!}+frac{x^8}{8!}-frac{x^{10}}{10!}+...

Find the Maclaurin series for $f\left(x\right)={x}^{4}\mathrm{cos}x$. Also, find the third term for this series.
The Maclaurin series for $\mathrm{cos}\left(x\right)$ is given by:
$\mathrm{cos}\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{{x}^{2n}}{\left(2n\right)!}=1-\frac{{x}^{2}}{2!}+\frac{{x}^{4}}{4!}-\frac{{x}^{6}}{6!}+\frac{{x}^{8}}{8!}-\frac{{x}^{10}}{10!}+...$
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The Maclaurin series for $\mathrm{cos}\left(x\right)$ is given by:
$\mathrm{cos}\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{{x}^{2n}}{\left(2n\right)!}=1-\frac{{x}^{2}}{2!}+\frac{{x}^{4}}{4!}-\frac{{x}^{6}}{6!}+\frac{{x}^{8}}{8!}-\frac{{x}^{10}}{10!}+...$
Multiply the Maclaurin series for and obtain the Maclaurin series for $f\left(x\right)={x}^{4}\mathrm{cos}\left(x\right)$ as shown below.
${x}^{4}\mathrm{cos}\left(x\right)={x}^{4}\left(\sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{{x}^{2n}}{\left(2n\right)!}\right)$
$=\sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{{x}^{4}\cdot {x}^{2n}}{\left(2n\right)!}$
$=\sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{{x}^{2n+4}}{\left(2n\right)!}$
Thus, the Maclaurin series for $f\left(x\right)={x}^{4}\mathrm{cos}\left(x\right)$ is,
$f\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{{x}^{2n+4}}{\left(2n\right)!}$
Now find the third term for this series as shown below.
$f\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{{x}^{2n+4}}{\left(2n\right)!}$
$={x}^{4}-\frac{{x}^{6}}{2!}+\frac{{x}^{8}}{4!}-\frac{{x}^{10}}{6!}+\frac{{x}^{12}}{8!}-...$
$={x}^{4}-\frac{{x}^{6}}{2}+\frac{{x}^{8}}{24}-\frac{{x}^{10}}{720}+\frac{{x}^{12}}{40320}-...$
Hence, the third term of the series is $\frac{{x}^{8}}{24}$

Jeffrey Jordon