Answer to Question #280123 in Real Analysis for abc

Question

For the function tan^(−1) 𝑥  find the infinite Taylor series at 𝑎 = 0, the radius of convergence, range of convergence, derivative, integral and the product with itself.  

Answer 1

$\text{Recall well-known series}, \\ \frac{1}{1-x} =1+x+x^{2} +\cdots =\sum _{n=0}^{\infty }x^{n} , \left|x\right|<1 \\ \begin{array}{l} {\Rightarrow \tan ^{-1} \left(x\right)=\int \frac{1}{1+x^{2} } dx =\int \frac{1}{1-\left(-x^{2} \right)} dx } \\ {=\int \sum _{n=0}^{\infty }\left(-x^{2} \right)^{n} dx ,\left|x^{2} \right|<1\Rightarrow \left|x\right|^{2} <1\Rightarrow \left|x\right|<1} \\ {=\int \sum _{n=0}^{\infty }\left(-1\right)^{n} x^{2n} dx ,\left|x\right|<1} \\ {=\sum _{n=0}^{\infty }\frac{\left(-1\right)^{n} x^{2n+1} }{2n+1} ,\left|x\right|<1} \end{array} \\ \Rightarrow R=1 \\ \text{When } x=-1,\sum _{n=0}^{\infty }\frac{\left(-1\right)^{n} x^{2n+1} }{2n+1} =\sum _{n=0}^{\infty }\frac{\left(-1\right)^{n+1} }{2n+1} \text{which converges by Alternating series test as } \frac{1}{2n+1} \to 0 \text{ and it also decreasing} \\ \text{When } x=1,\sum _{n=0}^{\infty }\frac{\left(-1\right)^{n} x^{2n+1} }{2n+1} =\sum _{n=0}^{\infty }\frac{\left(-1\right)^{n} }{2n+1} \text{which converges by Alternating series test as } \frac{1}{2n+1} \to 0 \text{ and it also decreasing} \\ \text{Hence, the Radius of convergence } R=1 \\ \text{The Range of convergence } I=[-1,1] \\ \text{The derivative } \frac{d}{dx} \left(\tan ^{-1} x\right)=\frac{1}{1+x^{2} } \\ \text{ To obtain the Integral, we shall use Integration by part, } \\ \text{Let } J=\int \tan ^{-1} \left(x\right)dx , u=\tan ^{-1} \left(x\right) \quad dv=dx \\ \hspace{98pt} du=\frac{1}{1+x^{2} } dx \qquad v=x \\ \Rightarrow J=\int \tan ^{-1} \left(x\right)dx =x\tan ^{-1} \left(x\right)-\frac{1}{2} \int \frac{2x}{1+x^{2} } dx \\ \hspace{95pt}=x\tan ^{-1} \left(x\right)-\frac{\ln \left(x^{2} +1\right)}{2} +C \\ \text{The Product} \\ {\left( {{{\tan }^{ - 1}}x} \right)^2} = {\left( {x - \frac{1}{3}{x^3} + \frac{1}{5}{x^5} + \cdots } \right)^2} = {x^2} - \frac{2}{3}{x^4} + \frac{{23}}{{45}}{x^6} - \cdots$