# 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})$$ Consider the following series: $$\sum_{n=1}^\infty(-1)^n\frac{1}{n}$$
Use the Alternating Series Test to show this series converges. Find the Maclaurin series for using the definition of a Maclaurin series. [Assume that has a power series expansion.Do not show that Rn(x) tends to 0.] Also find the associated radius of convergence. $$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}))$$ Which of the series, and which diverge?
Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series’ convergence or divergence.)
$$\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]}.$$