idku one year ago tell me if I am doing this correctly. I haven't touched math for a while now.

1. idku

(I am just reviewing some concepts from the past) I want to find the power series representation for $$\displaystyle\large\color{black}{\tan^{-1}(x)}$$ now I will show my work....

2. idku

$$\displaystyle\large\color{black}{\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n}$$ $$\displaystyle\large\color{black}{\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n}$$ $$\displaystyle\large\color{black}{\frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^nx^{2n}}$$ $$\displaystyle\large\color{black}{\int \frac{1}{1+x^2}~dx~=~\int~\sum_{n=0}^{\infty}(-1)^nx^{2n}~dx}$$ $$\displaystyle\large\color{black}{\tan^{-1}(x)~=~\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{2n+1}}$$

3. idku

and I am not placing +C, because I am looking for specific function - representation and not a family of functions.

4. idku

oh my power is off

5. idku

$$\displaystyle\large\color{black}{\tan^{-1}(x)~=~\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}}$$

6. idku

this should be correct, if I didn't err anywhere once again.

7. geerky42

What's the problem here? You already represented $$\tan^{-1}(x)$$ as series in your other question here.

8. geerky42
9. idku

yeah forgot about that..... my apologies.

10. idku

tnx for the confirmation, and bye:)

11. geerky42

Ok no problem.