How do you find the power series for $f(x)=int arctan(t^3)dt$ from [0,x] and determine its radius of convergence?

Question

2478 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-06T00:48:10

$int_0^x arctant^3dt = sum_(n=0)^oo (-1)^n x^(6n+4)/((6n+4)(2n+1))$

with radius of convergence $R=1$

Start from:

$arctanx = int_0^x (dt)/(1+t^2)$

Now the integrand function is the sum of a geometric series of ratio $-t^2$ :

$1/(1+t^2) = sum_(n=0)^oo (-1)^n(t^2)^n = sum_(n=0)^oo (-1)^nt^(2n)$

so:

$arctan x = int_0^x sum_(n=0)^oo (-1)^nt^(2n)$

This series has radius of convergence $R=1$ , so in the interval $x in (-1,1)$ we can integrate term by term:

$arctan x = sum_(n=0)^oo int_0^x (-1)^nt^(2n)dt = sum_(n=0)^oo (-1)^n x^(2n+1)/(2n+1)$

and the resulting series has the same radius of convergence.

Now substitute $x= t^3$ . the radius of convergence does not change since $abs x < 1 => abs(t^3) < 1$ :

$arctan t^3 = sum_(n=0)^oo (-1)^n (t^3)^(2n+1)/(2n+1) = sum_(n=0)^oo (-1)^n t^(6n+3)/(2n+1)$

and integrate again term by term:

$int_0^x arctant^3dt = sum_(n=0)^oo (-1)^n int_0^x t^(6n+3)/(2n+1)dt = sum_(n=0)^oo (-1)^n x^(6n+4)/((6n+4)(2n+1))$

How do you find the power series for f(x)=int arctan(t^3)dtf(x)=int arctan(t^3)dt from [0,x] and determine its radius of convergence?