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This question has come up twice in different tests and the instructions always point out that it should be solved using a graphic calculator. Fair enough, the answer is ≈ 1.76322...(goes on forever?).

But how do you approach $e^\frac1x = x$ analytically for that solution? Is there a way?

Jack Stout
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  • I doubt that the solution can be expressed as a combination of "elementary" functions.. – Peter Franek May 20 '15 at 11:11
  • The solution must be transcendental because of the known fact that $e^x$ is transcendental for every algebraic $x\ne 0$. So, the solution is not a root of a polynomial with integer coefficients. In particular, the decimal expansion neither terminates nor has a period. – Peter May 20 '15 at 11:24

6 Answers6

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The equation is:

$$e^{\frac 1x}=x$$

Raise everything to the $x$ power:

$$e=x^x$$

Now using super-square root in terms of Lambert's function leads us to:

$$x=\sqrt e_s =e^{W(1)}=\frac 1{W(1)}$$

AlienRem
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Let $x=\frac{1}{u}$. Then \begin{eqnarray} e^{u} &=& \ln \left( \frac{1}{u} \right) \\ \implies e^{u} &=& \ln1 - \ln u \\ &=& - \ln u \\ \implies \ln u + e^{u} &=& 0 \\ \iff e^{\ln u + e^{u}} &=& 1 \\ \iff e^{\ln u}.e^{e^{u}} &=& 1 \\ \iff ue^{e^{u}}=1 \end{eqnarray} I suspect the analytic solutions can be found through the Lambert W Function

Autolatry
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That's $f(\frac{1}{x})=1$, with Lambert W, and not extremely nice.

But it also formulates itself as a fixed point problem:

enter image description here

(computed here.)

Above we have $(g\circ g \circ \dots\circ g)(1)$, i.e. the value after 11 iterations of $g(x):=e^{\frac{1}{x}}$ applied to the solution guess $x=1$.

Nikolaj-K
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Now note that $e^{\frac{1}{x}}>0$ for $x<0$ and note that $x<0$ for $x<0$. Hence we only have to consider $x>0$.

If $e^{\frac{1}{x}}=x$, then $1/x=\ln(x)$ for $x>0$ , which means that $x\ln(x)=1$.

Now write $x\ln(x)=\ln(x)e^{\ln(x)}=1$ and note that Lambert's W function says then that $\ln(x)=W(1)$, hence $x=e^{W(1)}$

Alwin
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  • Thank you for the information. Solving it is clearly outside the scope of my current mathematical knowledge but it's great to learn about it. – Jack Stout May 20 '15 at 12:01
  • @jack "Lambert W function" means that solving it is outside the scope of everyone's current mathematical knowledge. We don't know how to relate the solutions to anything we actually understand, so we make up a name for the solution of the equation so that we can discuss it. The name is "Lambert W function". – MJD May 20 '15 at 12:48
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The following PARI/GP-program finds the solution using newton's method :

? x=1.5;for(j=1,6,x=x-(exp(1/x)-x)/(-1/x^2*exp(1/x)-1);print(j,"   ",x))
1   1.739986998347306487376609711
2   1.763077568142750917999087271
3   1.763222828791787414311562820
4   1.763222834351896702080716104
5   1.763222834351896710225201777
6   1.763222834351896710225201777

Since the derivate of $f(x)=e^{\frac{1}{x}}-x$ is $f'(x)=-\frac{1}{x^2}e^{\frac{1}{x}}-1$ , which is obviously negative for all $x>0$, the soltuion is moreover unique because for $x<0$, $f(x)$ is obviously positive.

Peter
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Here are the steps $$ \exp\left(\frac1x\right)=x $$ $$ \frac1x \exp\left(\frac1x\right)=1 $$ $$ \frac1x=W\left(1\right) =\Omega$$ $$ x=\frac1{\Omega}\approx\frac1{0.56714329}=1.76322$$ Where $W(z)$ is the Lambert W Function and $\Omega$ is the Omega Constant.

k170
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