points details my notes scalcet9 11.9.a d a power series representation for f(x) = arccot(x). observe that…

Explanation:

Step1: Recall geometric - series formula

We know that $\frac{1}{1 + t}=\sum_{n = 0}^{\infty}(-t)^{n}=1 - t+t^{2}-t^{3}+\cdots$ for $|t|\lt1$. Let $t = x^{2}$, then $\frac{1}{1 + x^{2}}=\sum_{n = 0}^{\infty}(-x^{2})^{n}=1 - x^{2}+x^{4}-x^{6}+\cdots$ for $|x|\lt1$.

Step2: Integrate the series

Since $\text{arccot}(x)=\int\frac{- 1}{1 + x^{2}}dx$, and $\frac{-1}{1 + x^{2}}=-(1 - x^{2}+x^{4}-x^{6}+\cdots)$. Integrating term - by - term, we have $\int-(1 - x^{2}+x^{4}-x^{6}+\cdots)dx=C - x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}-\cdots$.

Step3: Determine the constant C

We substitute $x = 0$ into $\text{arccot}(x)=C - x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}-\cdots$. Since $\text{arccot}(0)=\frac{\pi}{2}$, we get $C=\frac{\pi}{2}$.

Answer:

$\text{arccot}(x)=\frac{\pi}{2}-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}-\cdots=\frac{\pi}{2}+\sum_{n = 1}^{\infty}\frac{(-1)^{n}x^{2n + 1}}{2n+1},|x|\lt1$