I just finished my exam a few hours a go, and there was 1 question I couldn't answer. I was asked to derive the Taylor series of $\arg\tanh(x)$ using the fact that $$\tanh(x)=\frac{\sinh(x)}{\cosh(x)},$$ $\cosh(x)=\frac{e^x+e^{-x}}{2}$ and $\sinh(x)=\frac{e^x-e^{-x}}{2}$.
Now I'm still scratching my head trying to figure out how I should have done this. What I did is differentiate $\cosh(x)$ and $\sinh(x)$, but then I didn't know what to do next.

I don't know if they will give solutions to this exam, but if they do, it's at least in 2 to 3 months, and that's a long time. Also I searched on the internet but didn't find something useful (maybe I didn't search well enough??),

Thanks for your help !

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    $\begingroup$ Can't you simply perform a termwise integration of the Maclaurin series of $\frac{1}{1-x^2}=\frac{d}{dx}\text{arctanh}(x)$, which is a geometric series? $\endgroup$ – Jack D'Aurizio Aug 13 '18 at 14:44
  • $\begingroup$ Well, I didn't really think about it, but I guess it should work. But then, why do they give us $cosh(x)=\frac{e^x+e^{-x}}{2}$ and $sinh(x)=\frac{e^x-e^{-x}}{2}$ ? $\endgroup$ – Poujh Aug 13 '18 at 14:49
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    $\begingroup$ In order to figure out that $\frac{d}{dx}\text{arctanh}(x)=\frac{1}{1-x^2}$, probably. $\endgroup$ – Jack D'Aurizio Aug 13 '18 at 14:50
  • $\begingroup$ I didn't think it could be that simple. Thanks for your help ! $\endgroup$ – Poujh Aug 13 '18 at 14:56

A convenient way to arrange the computation is by differentiating


(with hopefully obvious shorthands) giving



$$t''c+2t's+tc=s, \\t'''c+3t''s+3t'c+ts=c, \\t''''c+4t'''s+6t''c+4t's+tc=s, \\t'''''c+5t''''s+10t'''c+10t''s+5t'c+ts=c, \\\cdots$$

If we set $x=0$, these simplify to

$$t_0=0, \\t'_0=1, \\t''_0+t_0=0, \\t'''_0+3t'_0=1, \\t''''_0+6t''_0+t_0=0, \\t'''''_0+10t'''_0+5t'_0=1, \\\cdots$$

More generally, the even coefficients are zero and

$$t_0^{(2n+1)}=1-\sum_{k=1}^{n} \binom {2n+1}{2k+1}t_0^{(2n+1-2k)}.$$

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