another question

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

Question

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

misty1212 · Answer

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

idku · Answer

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

idku · Answer

right ?

misty1212 · Answer

you got this !

idku · Answer

oh, I forgot something

idku · Answer

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

idku · Answer

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

idku · Answer

like this... correct ?

phi · Answer

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

idku · Answer

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

phi · Answer

yes

idku · Answer

tnx