Power Series Representation and finding radius of convergence.

[LukasKL]

Hi, I've been stuck on this problem for a good 30 minutes.

Find a power representation for the function and determine the radius of convergence.
f(x)=x^(2)arctan(x^(3))

I differentiated arctan(x) which is 1/(1+x^(2)) then changed it into the 1/(1-r) form so it becomes 1/(1-(-x^2)). I integrated both sides so it becomes arctan(x)=sigma (-1)^n (x^(2n+1))/(2n+1).

I am not sure what to do after this.

 

[HallsofIvy]

Hi, I've been stuck on this problem for a good 30 minutes.

Only 30 minutes? Why, you have hardly started!

Find a power representation for the function and determine the radius of convergence.
f(x)=x^(2)arctan(x^(3))

I differentiated arctan(x) which is 1/(1+x^(2)) then changed it into the 1/(1-r) form so it becomes 1/(1-(-x^2)). I integrated both sides so it becomes arctan(x)=sigma (-1)^n (x^(2n+1))/(2n+1).

I am not sure what to do after this.

Good! Now replace x by \(\displaystyle x^3\) to get \(\displaystyle arctan(x^3)= \sum (-1)^n ((x^3)^{2n+1})/(2n+1)= \sum (-1)^n x^{6n+3}/(2n+1)\)
Then multiply by \(\displaystyle x^2\) to get \(\displaystyle x^2arctan(x^3)= \sum (-1)^n x^{6n+5}/(2n+1)\)

Use the ratio test to find the radius of convergence.