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

Question

idku · Answer

(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....

idku · Answer

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

idku · Answer

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

idku · Answer

oh my power is off

idku · Answer

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

idku · Answer

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

geerky42 · Answer

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

geerky42 · Answer

http://openstudy.com/users/idku#/updates/554a6664e4b048e824ff3507

idku · Answer

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

idku · Answer

tnx for the confirmation, and bye:)

geerky42 · Answer

Ok no problem.