How do you find a power series representation for #x^2 arctan(x^3# and what is the radius of convergence?

Question

Luke Alvarez · Accepted Answer

#x^2arctan(x^3)=sum_(n=0)^oo(-1)^n/(2n+1)x^(6n+5)=x^5-x^11/3+x^17/5+...#

The radius of convergence is #R=1#.

Start with the well known Maclaurin series: #1/(1-x)=sum_(n=0)^oox^n# Replacing #x# with #-x^2#: #1/(1+x^2)=sum_(n=0)^oo(-x^2)^n=sum_(n=0)^oo(-1)^nx^(2n)# Integrating to get #arctan(x)#: #arctan(x)=intdx/(1+x^2)=sum_(n=0)^oo(-1)^nintx^(2n)dx# #arctan(x)=C+sum_(n=0)^oo(-1)^nx^(2n+1)/(2n+1)# Letting #x=0# shows #C=0#: #arctan(x)=sum_(n=0)^oo(-1)^n/(2n+1)x^(2n+1)# Replacing #x# with #x^3#: #arctan(x^3)=sum_(n=0)^oo(-1)^n/(2n+1)(x^3)^(2n+1)=sum_(n=0)^oo(-1)^n/(2n+1)x^(6n+3)# Multiplying by #x^2#, which can be brought into the series: #x^2arctan(x^3)=x^2sum_(n=0)^oo(-1)^n/(2n+1)x^(6n+3)=color(blue)(sum_(n=0)^oo(-1)^n/(2n+1)x^(6n+5)# This is the Maclaurin series for #x^2arctan(x^3)#. To find its radius of convergence, find the ratio #abs(a_(n+1)/a_n)# where #a_n=(-1)^n/(2n+1)x^(6n+5)#: #abs(a_(n+1)/a_n)=abs(((-1)^(n+1)/(2(n+1)+1)x^(6(n+1)+5))/((-1)^n/(2n+1)x^(6n+5)))=abs(((-1)^(n+1)/(2n+3)x^(6n+11))/((-1)^n/(2n+1)x^(6n+5)))# #abs(a_(n+1)/a_n)=abs((-1)^(n+1)/(-1)^n((2n+1)/(2n+3))x^(6n+11)/x^(6n+5))=abs(x^6((2n+1)/(2n+3)))# Through the ratio test, we know the series converges when #lim_(nrarroo)abs(a_(n+1)/a_n)<1#. Finding the limit: #lim_(nrarroo)abs(a_(n+1)/a_n)=lim_(nrarroo)abs(x^6((2n+1)/(2n+3)))=absx^6lim_(nrarroo)abs((2n+1)/(2n+3))# The internal limit is #1#. We want this when it's less than #1#: #absx^6<1" "=>" "absx<1# Thus the radius of convergence is #color(blue)(R=1#.

Axel Alvarez · Answer

undefined To find a power series representation for $x^2 \arctan(x^3)$, we can use the formula for the power series expansion of $\arctan(x)$:

$$ \arctan(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1} $$

First, replace $x$ with $x^3$ in the series formula:

$$ \arctan(x^3) = \sum_{n=0}^{\infty} (-1)^n \frac{(x^3)^{2n+1}}{2n+1} $$

$$ = \sum_{n=0}^{\infty} (-1)^n \frac{x^{6n+3}}{2n+1} $$

Then, multiply by $x^2$:

$$ x^2 \arctan(x^3) = x^2 \sum_{n=0}^{\infty} (-1)^n \frac{x^{6n+3}}{2n+1} $$

$$ = \sum_{n=0}^{\infty} (-1)^n \frac{x^{6n+5}}{2n+1} $$

So, the power series representation for $x^2 \arctan(x^3)$ is:

$$ x^2 \arctan(x^3) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{6n+5}}{2n+1} $$

The radius of convergence of this series can be found using the ratio test. Let $a_n = (-1)^n \frac{1}{2n+1} x^{6n+5}$. Applying the ratio test:

$$ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(-1)^{n+1} \frac{1}{2(n+1)+1} x^{6(n+1)+5}}{(-1)^n \frac{1}{2n+1} x^{6n+5}} \right| $$

$$ = \lim_{n \to \infty} \left| \frac{-1}{2n+3} x^{6n+11} \cdot \frac{2n+1}{1} x^{-6n-5} \right| $$

$$ = \lim_{n \to \infty} \left| -\frac{x^6}{2n+3} \right| $$

Since $\lim_{n \to \infty} \frac{x^6}{2n+3} = 0$ for all $x$, the radius of convergence is infinite, which means the series converges for all real numbers $x$.