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$3f(x)=e^{x}+e^{\alpha x}+e^{\alpha^2 x}$ 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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    @joriki : Yes correct. Now your formula works fine. If we combine both relation. Can be gotten super thing. $f(3x)=1+9f(x)f'(x)f''(x)$ Thanks2012-08-10

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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.