Power Series and Functions — Question 5

*Question 5 Derive the Maclaurin series for $\arctan x$ by integrating the series for $1/(1+x^2)$. Explain the odd powers and determine the endpoint behavior.

*Question 5 – Solution

With geometric ratio $-t^2$, $$\frac1{1+t^2}=\sum_{n=0}^{\infty}(-1)^n t^{2n},\qquad |t|<1.$$

$$\arctan x=\int_0^x\frac{dt}{1+t^2}=\boxed{\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1}},\qquad |x|<1.$$ Integrating the even powers $t^{2n}$ produces only odd powers $x^{2n+1}$; the value at $0$ fixes the constant.

At $x=1$, the alternating series converges to $\pi/4$. At $x=-1$, it converges to $-\pi/4$. Both are conditional.

The radius is $R=1$ and the interval is $\boxed{[-1,1]}$.