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What is the Taylor series of $$\ln \frac{1-x^2}{1+x^2}$$ ?

I started by evaluating the first derivatives but the more I go, the more complicated they are and I can't identify a pattern.

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  • $\begingroup$ It's better to parse it on two parts. $\endgroup$ – openspace Jan 20 '16 at 10:13
  • $\begingroup$ What is the centre of the Taylor series that you desire? $\endgroup$ – Alan Wang Jan 20 '16 at 10:13
  • $\begingroup$ @AlanWang, around $x=0$. $\endgroup$ – Elimination Jan 20 '16 at 10:15
  • $\begingroup$ @openspace, that's a good idea actually. $\endgroup$ – Elimination Jan 20 '16 at 10:15
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    $\begingroup$ In your book you will find expansion for $log(1+x)$. Replace $x$ by $\pm x^2$ and use properties of $\ln$. $\endgroup$ – Maesumi Jan 20 '16 at 10:22
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Hint 1 consider $$ln(1-x^{2})+ln(1+x^{2})$$ Hint 2 consider $$(1-x^{2}) = (1-x)(1+x)$$ and $$1+x^{2}= (1-ix)(1+ix)$$

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  • $\begingroup$ It should be $\ln(1-x^2)-\ln(1+x^2)$ for the quotient, and you can use the logarithm series directly from this form, no further factorization required. $\endgroup$ – LutzL Jan 20 '16 at 11:03
  • $\begingroup$ @LutzL yes, miss the "-" $\endgroup$ – openspace Jan 20 '16 at 11:05
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As $\log\dfrac ab=\log a-\log b$

$$\ln\dfrac{1-x^2}{1+x^2}=\ln(1-x^2)-\ln(1+x^2)$$

Now use $\ln(1-u)=-\sum_{r=0}^\infty\dfrac{u^r}r$ as $-1\le u<1$

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The derivative of $\log\frac{1-x^2}{1+x^2}$ is: $$ \frac{4x}{x^4-1} = -4x-4x^5-4x^9-\ldots = -4\sum_{n\geq 0}x^{1+4n} \tag{1}$$ hence it follows that: $$ \log\left(\frac{1-x^2}{1+x^2}\right) = -4\sum_{n\geq 0}\frac{x^{4n+2}}{4n+2} = -2\sum_{n\geq 0}\frac{x^{4n+2}}{2n+1}.\tag{2}$$ You may also notice that $\log\frac{1-x^2}{1+x^2}=-2\,\text{arctanh}(x^2)$.

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  • $\begingroup$ Very nice solution! $\endgroup$ – Elimination Jan 20 '16 at 15:46

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