# To find the closed form of $f^{-1}(x)$ if $3f(x)=e^{x}+e^{\alpha x}+e^{\alpha^2 x}$

$$3f(x)=e^{x}+e^{\alpha x}+e^{\alpha^2 x}$$ 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.

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Do you have any reason to think that a closed-form solution exists? It seems to me to be very unlikely. – Robert Israel Aug 9 '12 at 21:34
@RobertIsrael : We know we can find a closed form $y''=y$ and $y(x)=\cosh x=(e^x+e^{-x})/2$ and $y^{-1}(x)=arccosh x=\int \frac{dx}{\sqrt{x^2-1}}$ .I wanted to go one more step and I wrote $y'''=y$. – Mathlover Aug 9 '12 at 21:44
I was thinking whether it might be possible to do something like $$\arcsin'=\frac1{\sin'}=\frac1{\sqrt{1-\sin^2}}=\frac1{\sqrt{1-x^2}}\;,$$ which relies on $\sin^2+\sin'^2=1$. The best I came up with is $f(x)^3+f'(x)^3+f''(x)^3=18+3f(3x)$, which isn't good enough for the purpose, but still sort of interesting. – joriki Aug 9 '12 at 22:08
Have you looked into applying Lagrangian inversion to the series you have? – J. M. Aug 10 '12 at 0:59
@joriki : Addition to this,$f(x)^3+f′(x)^3+f′′(x)^3=1+3f(x)f′(x)f′′(x)$ It is easy to proof after applying derivative both side and to know the fact $f(0)=1$,$f'(0)=0$ $f''(0)=0$ – Mathlover Aug 10 '12 at 8:53