Question

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

Answer 1

`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`.