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

1 Answer
Mar 17, 2017

#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#.

Explanation:

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#.