Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

$$f(x)=f(\alpha x)=f(\alpha^2 x)=\frac{e^{x}+2e^{-\frac{x}{2}} \cos{\frac{x\sqrt{3}}{2}}}{3}=$$



$\alpha=e^{\frac{2\pi i}{3} }=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$

My first attempt to find $f^{-1}(x)$:

$$3x=e^{f^{-1}(x)}+e^{\alpha f^{-1}(x)}+e^{\alpha^2 f^{-1}(x)}$$


$p+p^{\alpha }+p^{\alpha^2 }=3x$

$$p'(p+(-\frac{1}{2}+i\frac{\sqrt{3}}{2}) p^{\alpha}+(-\frac{1}{2}-i\frac{\sqrt{3}}{2})p^{\alpha^2 })=3p$$

$$p'(p+(-\frac{1}{2}+i\frac{\sqrt{3}}{2}) p^{\alpha}+(-\frac{1}{2}-i\frac{\sqrt{3}}{2})p^{\alpha^2 })=3p$$

$$p'(p -\frac{1}{2}(p^{\alpha }+p^{\alpha^2 })+i\frac{\sqrt{3}}{2} (p^{\alpha}-p^{\alpha^2 })=3p$$

$$i\frac{\sqrt{3}}{2} (p^{\alpha}-p^{\alpha^2 })=\frac{3p}{p'}+\frac{3x}{2}-\frac{3p}{2}$$

$$-\frac{3}{4} (p^{2\alpha}+p^{2\alpha^2 }-2p^{-1})=(\frac{3p}{p'}+\frac{3x}{2}-\frac{3p}{2})^2$$

$-\frac{3}{4} ((3x-p)^2-4p^{-1})=(\frac{3p}{p'}+\frac{3x}{2}-\frac{3p}{2})^2$

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 $g(x)=f^{-1}(x)$










$$u u'^2+u^2u''=x$$

if $u=z^{1/2}$ then


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.

share|cite|improve this question
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
up vote 1 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.