Question

another question


power series rep. for tan^(-1)x

Answer 1

differentiate get
\[\frac{1}{1+x^2}\] find the power series, integrate term by term

Answer 2

I know that
\[\frac{d}{dx}\left(\tan^{-1}x\right)=\frac{1}{1+x^2}\]
(this can be proven

tan^(-1)x=y
tan(y)=x

derivative,

dy/dx * sec^2(y) = 1
dy/dx * ( tan^2(y) + 1) = 1
dy/dx = 1/(1+tan^2y)
dy/dx = 1/(1+x^2)

the prove is here, that is no the concern)

BUT,

\[\frac{1}{1-(-x^2)}=\sum_{n=1}^{\infty}x^{2n}\]\[\tan^{-1}(x)=\sum_{n=1}^{\infty}\frac{x^{2n+1}}{2n+1}\]

Answer 3

right ?

Answer 4

you got this !

Answer 5

oh, I forgot something

Answer 6

\[\frac{1}{1-(-x^2)}=\sum_{n=1}^{\infty}(-1)^{n}x^{2n}\]

Answer 7

\[\tan^{-1}x=\sum_{n=1}^{\infty}(-1)^{n}\frac{x^{2n+1}}{2n+1}\]

Answer 8

like this... correct ?

Answer 9

again, the lower limit is n=0 not n=1

Answer 10

oh, so all the work is correct besides that it is n=0 (and not n=1) ?

Answer 11

yes

Answer 12

tnx