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!}+...

Question
Series
asked 2021-02-02
Find the Maclaurin series for \(f(x)=x^4\cos x\). Also, find the third term for this series.
The Maclaurin series for \(\cos(x)\) is given by:
\(\cos(x)=\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\frac{x^{10}}{10!}+...\)

Answers (1)

2021-02-03
The Maclaurin series for \(\cos(x)\) is given by:
\(\cos(x)=\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\frac{x^{10}}{10!}+...\)
Multiply the Maclaurin series for \(\cos(x)\text{ by }x^4\) and obtain the Maclaurin series for \(f(x)=x^4\cos(x)\) as shown below.
\(x^4\cos(x)=x^4(\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!})\)
\(=\sum_{n=0}^\infty(-1)^n\frac{x^4\cdot x^{2n}}{(2n)!}\)
\(=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+4}}{(2n)!}\)
Thus, the Maclaurin series for \(f(x)=x^4\cos(x)\) is,
\(f(x)=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+4}}{(2n)!}\)
Now find the third term for this series as shown below.
\(f(x)=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+4}}{(2n)!}\)
\(=x^4-\frac{x^6}{2!}+\frac{x^8}{4!}-\frac{x^{10}}{6!}+\frac{x^{12}}{8!}-...\)
\(=x^4-\frac{x^6}{2}+\frac{x^8}{24}-\frac{x^{10}}{720}+\frac{x^{12}}{40320}-...\)
Hence, the third term of the series is \(\frac{x^8}{24}ZS
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