find a power series representation for f(x) = arccot(x). solution we observe that f(x) = -1/(1 + x^2) and…

Explanation:

Step1: Recall geometric - series formula

We know that $\frac{1}{1 + t}=\sum_{n = 0}^{\infty}(-t)^{n}$, for $|t|\lt1$. Let $t=x^{2}$, then $\frac{-1}{1 + x^{2}}=-\sum_{n = 0}^{\infty}(-x^{2})^{n}=-\sum_{n = 0}^{\infty}(- 1)^{n}x^{2n}=\sum_{n = 0}^{\infty}(-1)^{n + 1}x^{2n}$, for $|x^{2}|\lt1$ (i.e., $|x|\lt1$).

Step2: Integrate the power - series

$\text{arccot}(x)=\int\frac{-1}{1 + x^{2}}dx=\int\sum_{n = 0}^{\infty}(-1)^{n+1}x^{2n}dx$. By the power - rule of integration $\int x^{m}dx=\frac{x^{m + 1}}{m+1}+C$ ($m\neq - 1$), we have $\text{arccot}(x)=C+\sum_{n = 0}^{\infty}\frac{(-1)^{n + 1}}{2n+1}x^{2n + 1}$.

Step3: Determine the constant $C$

When $x = 0$, $\text{arccot}(0)=\frac{\pi}{2}$, so $C=\frac{\pi}{2}$. Then $\text{arccot}(x)=\frac{\pi}{2}-\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}x^{2n+1}$.

Step4: Find the radius of convergence

The power - series for $\frac{-1}{1 + x^{2}}=\sum_{n = 0}^{\infty}(-1)^{n + 1}x^{2n}$ converges for $|x^{2}|\lt1$, i.e., $|x|\lt1$. When we integrate a power - series $\sum_{n = 0}^{\infty}a_{n}x^{n}$ with radius of convergence $R$, the radius of convergence of the integrated power - series $\sum_{n = 0}^{\infty}\frac{a_{n}}{n + 1}x^{n+1}$ is the same as that of the original power - series. So the radius of convergence of the power - series for $\text{arccot}(x)$ is $R = 1$.

Answer:

$\text{arccot}(x)=\frac{\pi}{2}-\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{2n+1}x^{2n + 1}$, radius of convergence $R = 1$