TITLE: Expansion of $(\sinh(x))^{\frac15}$ around 0 for x > 0
QUESTION [0 upvotes]: I'm aware of the series expansion of the hyperbolic functions, but how does one expand a fractional power of sinus hyperbolicus, i.e. e.g. $(\sinh(x))^{\frac15}$?

REPLY [0 votes]: The way I usually approach these kind of expansions is by looking at the power series expansion of the inner and outer function. For this problem, the series expansion for $\sinh(x)$ (inner function) is given by
$$\sinh(x)=\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)!}=x+\frac{x^3}{6}+\frac{x^5}{120}+\mathcal{O}(x^7)$$
and for the outer function we have
$$(1+x)^{\frac{1}{5}}=1+\frac{x}{5}-\frac{2x^2}{25}+\mathcal{O}(x^3).$$
Therefore, only including the highest order, the expansion reads:
$$\sinh^{\frac{1}{5}}(x)=\Big(x+\frac{x^3}{6}+\mathcal{O}(x^5)\Big)^{\frac{1}{5}}=x^{\frac{1}{5}}\Big(1+\frac{x^2}{6}+\mathcal{O}(x^4)\Big)^{\frac{1}{5}}=x^{\frac{1}{5}}\Big(1+\frac{x^2}{30}+\mathcal{O}(x^4)\Big).$$
The next highest order term would have the form
\begin{align}
\sinh^{\frac{1}{5}}(x)&=\Big(x+\frac{x^3}{6}+\frac{x^5}{120}+\mathcal{O}(x^7)\Big)^{\frac{1}{5}}\\&=x^{\frac{1}{5}}\Big(1+\frac{x^2}{6}+\frac{x^4}{120}+\mathcal{O}(x^6)\Big)^{\frac{1}{5}}\\&=x^{\frac{1}{5}}\Big(1+\frac{1}{5}\Big(\frac{x^2}{6}+\frac{x^4}{120}\Big)-\frac{2}{25}\Big(\frac{x^2}{6}+\frac{x^4}{120}\Big)^2+\mathcal{O}(x^6)\Big)\\&=x^{\frac{1}{5}}\Big(1+\frac{x^2}{30}-\frac{x^4}{1800} +\mathcal{O}(x^6)\Big)
\end{align}
where in the last line because of the precision up to order $x^6$ the higher order terms from the expansion of $\big(\frac{x^2}{6}+\frac{x^4}{120}\big)^2$ were discarded. This procedure then can be iterated futher and further for higher orders in the expansion.
Bonus: For the more generalized version of your problem, the expansion up to the second highest order has the from
$$\sinh^\alpha(x)=x^\alpha\bigg(1+\frac{\alpha}{6}x^2+\frac{\alpha(5\alpha-2)}{360}x^4+\mathcal{O}(x^6)\bigg).$$
{"set_name": "stack_exchange", "score": 0, "question_id": 4031604}