Expansion of $(\sinh(x))^{\frac15}$ around 0 for x > 0

Question

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. $(\sinh(x))^{\frac15}$?

Answer 1

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 $$\sinh^\alpha(x)=x^\alpha\bigg(1+\frac{\alpha}{6}x^2+\frac{\alpha(5\alpha-2)}{360}x^4+\mathcal{O}(x^6)\bigg).$$

Answer 2

If you want a series expansion with integer power, let $x=t^5$ $$y=\sqrt[5]{\sinh \left(t^5\right)}\implies \log(y)=\frac 15 \log \left(\sinh \left(t^5\right)\right)$$ Now, compose series (one piece at the time) $$\sinh \left(t^5\right)=t^5+\frac{t^{15}}{6}+\frac{t^{25}}{120}+\frac{t^{35}}{5040}+\frac{t^{45}}{362880}+O\left(t^{55}\right)$$ $$\log \left(\sinh \left(t^5\right)\right)=5 \log (t)+\frac{t^{10}}{6}-\frac{t^{20}}{180}+\frac{t^{30}}{2835}-\frac{t^{40}}{37800}+ O\left(t^{50}\right)$$ $$\log(y)=\log (t)+\frac{t^{10}}{30}-\frac{t^{20}}{900}+\frac{t^{30}}{14175}-\frac{t^{40}}{18900 0}+O\left(t^{50}\right)$$ $$y=e^{\log(y)}=t+\frac{t^{11}}{30}-\frac{t^{21}}{1800}+\frac{t^{31}}{25200}-\frac{131 t^{41}}{45360000}+O\left(t^{51}\right)$$