I would be grateful if someone could check what I've worked out:

$$ f(x)=\arctan\left(\frac{3x+2}{3x-2}\right)\implies f'(x)=\frac{1}{1+(\frac{3x+2}{3x-2})^2}\cdot \frac{3(3x-2)-3(3x+2)}{(3x-2)^2}$$

$$=\frac{(3x-2)^2}{(3x-2)^2+3x+2)^2}\cdot \frac{-12}{(3x-2)^2}=-\frac{3}{2}\cdot \frac{1}{1+(\frac{3}{2}x)^2}$$

$$=-\frac{3}{2} \cdot \frac{1}{1+(\frac{3}{2}x)^2}=-\frac{3}{2} \sum_{k=0}^{\infty}\left(-\frac{3}{2}x^2\right)^k $$

Which implies $$f(x)=\int f'(x)dx=-\frac{3}{2} \sum_{k=0}^{\infty}\left(-\frac{3}{2} \frac{x^{2k+1}}{2k+1}\right)$$

Radius of convergence:

$$\Big|(\frac{3}{2}x)^2\Big|<1 \Rightarrow -\frac{2}{3}<x<\frac{2}{3}$$

Is this correct? Thank you in advance

  • 1
    $\begingroup$ The procedure is a good one except for a small detail. You need a definite integral. The constant term is not correct. I would prefer a dummy variable of integration other than $x$. This might have been enough to prevent the glitch. $\endgroup$ – André Nicolas Apr 23 '13 at 6:16
  • $\begingroup$ Yees, so can I assume that $C (=const) =f'(0)$ What gives me $\frac{3}{4} \cdot \pi$? $\endgroup$ – fdhd Apr 23 '13 at 6:19
  • $\begingroup$ @AndréNicolas I can't see how changing the variable helps with that. You either remember to add $C$ or you don't. How does the variable help? $\endgroup$ – Git Gud Apr 23 '13 at 6:22
  • $\begingroup$ Something like that. I think it might be $\frac{\pi}{6}$ (if you want a $-\frac{3}{2}$ in front) but my arithmetic is not good. $\endgroup$ – André Nicolas Apr 23 '13 at 6:23
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    $\begingroup$ You should add the constant of integration $C$ to your equation and then plug $x=0$ which gives you $C=f(0)$. $\endgroup$ – Mhenni Benghorbal Apr 23 '13 at 6:23

Here is what you are missing,

$$ f(x)=\int f'(x)dx=-\frac{3}{2} \sum_{k=0}^{\infty}\left((-1)^k\frac{3}{2} \frac{x^{2k+1}}{2k+1}\right) +C $$

$$ \implies f(0) = \arctan(-1) = 0 + C \implies C=-\frac{\pi}{4}. $$

  • $\begingroup$ Isn't $tan \frac{\pi}{4} = 1$? And then $C= \frac{3 \pi}{4}$? $\endgroup$ – fdhd Apr 23 '13 at 6:55
  • $\begingroup$ @fdhd: I forgot the minus sign, it should be $C=-\frac{\pi}{4}$ $\endgroup$ – Mhenni Benghorbal Apr 23 '13 at 8:15

You missed $(-1)^k\left(\frac{3}{2}\right)^{k}:$ $$f'(x)=-\frac{3}{2} \sum_{k=0}^{\infty}\left(-\frac{3}{2}x^2\right)^k=\frac{3}{2} \sum_{k=0}^{\infty}(-1)^{k+1}\left(\frac{3}{2}x^2\right)^k= \sum_{k=0}^{\infty}(-1)^{k+1}\left(\frac{3}{2}\right)^{k+1} x^2.$$ Then $$ f(x)=\int{f'(x)\,dx}= \sum_{k=0}^{\infty}(-1)^{k+1}\left(\frac{3}{2}\right)^{k+1} \frac{x^{2k+1}}{2k+1}+C.$$ From $f(0)=\arctan(-1)=-\dfrac{\pi}{4}$ we have $C=\dfrac{\pi}{4}.$


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