# 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
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:
$$\cos(x)=\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\frac{x^{10}}{10!}+...$$

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 ### Relevant Questions asked 2020-10-19 a) Find the Maclaurin series for the function \(f(x)=\frac11+x$$
b) Use differentiation of power series and the result of part a) to find the Maclaurin series for the function
$$g(x)=\frac{1}{(x+1)^2}$$
c) Use differentiation of power series and the result of part b) to find the Maclaurin series for the function
$$h(x)=\frac{1}{(x+1)^3}$$
d) Find the sum of the series
$$\sum_{n=3}^\infty \frac{n(n-1)}{2n}$$
This is a Taylor series problem, I understand parts a - c but I do not understand how to do part d where the answer is $$\frac72$$
Determine the radius of convergence of this series.
$$\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}$$
Find the values of x for which the given geometric series converges. Also, find the sum of the series (as a function of x) for those values of x.
a) Find the values of x for which the given geometric series converges.
b) Find the sum of the series
$$\sum_{n=0}^\infty(-\frac12)^n(x-5)^n$$
A) Does this series converge? If yes, towards what number?
B) Find the first 5 terms of the sequence of partial sums in this series.
C) What is the general term of this sequence of partial sums?
$$\sum_{k=1}^\infty(\frac{2}{k}-\frac{2}{k+1})$$
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Use the Alternating Series Test to show this series converges.
Find the Maclaurin series for using the definition of a Maclaurin series. [Assume that has a power series expansion.Do not show that Rn(x) tends to 0.] Also find the associated radius of convergence. $$f(x)=(1-x)^{-2}$$
Find a formula for the nth partial sum of the series and use it to determine whether the series converges or diverges. If a series converges, find its sum.
$$\sum_{n=1}^\infty(\cos^{-1}(\frac{1}{n+1})-\cos^{-1}(\frac{1}{n+2}))$$
$$\sum_{n=1}^\infty \frac{e^n}{1+e^{2n}}$$
$$\displaystyle{f{{\left({x}\right)}}}={5}{\cos{{\left({\frac{{\pi{x}}}{{{8}}}}\right)}}}$$
Let $$\displaystyle{S}_{{{N}}}{\left({x}\right)}={\frac{{{4}}}{{\pi}}}\ {\sum_{{{n}={1}}}^{{{N}}}}\ {\frac{{{1}\ -\ {\left(-{1}\right)}^{{{n}}}}}{{{n}^{{{3}}}}}}\ {\sin{{\left({n}{x}\right)}}}.$$
Construct graphs of $$\displaystyle{S}_{{{N}}}{\left({x}\right)}\ {\quad\text{and}\quad}\ {x}{\left(\pi\ -\ {x}\right)},\ {f}{\quad\text{or}\quad}\ {0}\ \leq\ {x}\ \leq\ \pi,\ {f}{\quad\text{or}\quad}\ {N}={2}\ {\quad\text{and}\quad}\ {t}{h}{e}{n}\ {N}={10}.$$
$$\displaystyle{\frac{{{4}}}{{\pi}}}\ {\sum_{{{n}={1}}}^{{\infty}}}\ {\frac{{{1}\ -\ {\left(-{1}\right)}^{{{n}}}}}{{{n}^{{{3}}}}}}\ {\sin{{\left({n}{x}\right)}}}\ {o}{n}\ {\left[{0},\ \pi\right]}.$$