$$3f(x)=e^{x}+e^{\alpha x}+e^{\alpha^2 x}$$ where $\alpha=e^{\frac{2\pi i}{3} }$
I would like to find a closed form of $ f^{-1}(x)$
$$f(x)=\sum \limits_{k=0}^\infty \frac{x^{3k}}{(3k)!}$$
We can see easily that $f'''(x)=f(x)$
$$f(x)=f(\alpha x)=f(\alpha^2 x)=\frac{e^{x}+2e^{-\frac{x}{2}} \cos{\frac{x\sqrt{3}}{2}}}{3}=$$
$\alpha^3=1$
$\alpha^2+\alpha+1=0$
$\alpha=e^{\frac{2\pi i}{3} }=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$
My first attempt to find $f^{-1}(x)$:
$$3x=e^{f^{-1}(x)}+e^{\alpha f^{-1}(x)}+e^{\alpha^2 f^{-1}(x)}$$
$$p(x)=e^{f^{-1}(x)}$$
$p+p^{\alpha }+p^{\alpha^2 }=3x$
$$p'(p+(-\frac{1}{2}+i\frac{\sqrt{3}}{2}) p^{\alpha}+(-\frac{1}{2}-i\frac{\sqrt{3}}{2})p^{\alpha^2 })=3p$$
$$p'(p+(-\frac{1}{2}+i\frac{\sqrt{3}}{2}) p^{\alpha}+(-\frac{1}{2}-i\frac{\sqrt{3}}{2})p^{\alpha^2 })=3p$$
$$p'(p -\frac{1}{2}(p^{\alpha }+p^{\alpha^2 })+i\frac{\sqrt{3}}{2} (p^{\alpha}-p^{\alpha^2 })=3p$$
$$i\frac{\sqrt{3}}{2} (p^{\alpha}-p^{\alpha^2 })=\frac{3p}{p'}+\frac{3x}{2}-\frac{3p}{2}$$
$$-\frac{3}{4} (p^{2\alpha}+p^{2\alpha^2 }-2p^{-1})=(\frac{3p}{p'}+\frac{3x}{2}-\frac{3p}{2})^2$$
$-\frac{3}{4} ((3x-p)^2-4p^{-1})=(\frac{3p}{p'}+\frac{3x}{2}-\frac{3p}{2})^2$
After here ,I am not sure that there is an easy solution. Maybe someone can give hint what to do for next step.
My second attempt to find $f^{-1}(x)$:
$f(g(x))=x$ ---> where $g(x)=f^{-1}(x)$
$$f'(g(x))g'(x)=1$$
$$f'(g(x))=\frac{1}{g'(x)}$$
$$f''(g(x))g'(x)=(\frac{1}{g'(x)})'$$
$$f''(g(x))=\frac{1}{g'(x)}(\frac{1}{g'(x)})'$$
$$f'''(g(x))g'(x)=(\frac{1}{g'(x)}(\frac{1}{g'(x)})')'$$
$$f(g(x))g'(x)=(\frac{1}{g'(x)}(\frac{1}{g'(x)})')'$$
$$f(g(x))=x=\frac{1}{g'(x)}(\frac{1}{g'(x)}(\frac{1}{g'(x)})')'$$
$$\frac{1}{g'(x)}=u(x)$$
$$u[uu']'=x$$
$$u u'^2+u^2u''=x$$
if $u=z^{1/2}$ then
$$z^{1/2}z''=2x$$
Here again, I do not know how to solve that differential equation. Any hint to solve it?
I also would like to share some interesting property of that function.
$9f^2(x)=(e^{x}+e^{\alpha x}+e^{\alpha^2 x})^2=e^{2x}+e^{\alpha 2x}+e^{\alpha^2 2x}+2(e^{-x}+e^{-\alpha x}+e^{-\alpha^2 x})$
$$3f^2(x)=f(2x)+2f(-x)$$ $$f(2x)=3f^2(x)-2f(-x)$$
Could you please advice a method to find $f^{-1}(x)$ in closed form such as integral expression of elementary functions. (Actually, I am looking for an expression that it is similiar to $\arcsin(x)=\int\frac{1}{\sqrt{1-x^2}}dx$, if possible)
Thank you for hints and for answers.
If you are really interested in solving exponential equations in "elementary functions", I once posted an answer to a similar problem on: AoPS. You may try to adjust that argument to your case.