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$$n=\frac{e^x}{\log x}$$

I've been thinking for several hours about how to find the inverse function, but I always get to a nested function.

They can help me or recommend some literature that allows me to solve this problem. In theory there is its inverse clearly not with elementary functions, I am interested in knowing and learning about new techniques for this type of functions. I appreciate the help.

Estudiante Inversa
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    you can use: https://en.wikipedia.org/wiki/Lagrange_inversion_theorem to acquire a series representation of the inverse – Sidharth Ghoshal Nov 28 '21 at 23:29
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    Using successive approximation, you can obtain an asymptotic expansion for large $n$: $$ x \sim \log n + \log \log \log n + \frac{{\log \log \log n}}{{(\log n)(\log \log n)}} + \cdots $$ This gives you one of the solutions asymptotically. – Gary Nov 28 '21 at 23:38
  • @Gary Thanks can you give me a link where I can look at this. – Estudiante Inversa Nov 28 '21 at 23:40
  • I cannot really give you a link. If you take the logarithm of your expression and re-order, you find $x=\log n+\log\log x$. Now you substitute $x$ on the right-hand side by the whole right-hand side and do some re-expansions using things like $\log(1+w)=w-\ldots$. – Gary Nov 28 '21 at 23:43
  • @Gary Thank you!! – Estudiante Inversa Nov 28 '21 at 23:45
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    Note that for $n = e^{e^{W(1)}}/W(1) \approx 10.281705230887$ there is one real solution of $x=e^{W(1)}\approx 1.76322283435$ using the Lambert W-function and $W(1)\approx 0.56714329$; for smaller $n$ there are no real solutions and for larger $n$ there are two real solutions – Henry Nov 28 '21 at 23:56

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Inverse function isn't the right term here, because you have an equation, not a function. A function can have partial inverses, only bijective functions have an inverse function, and an equation can have solutions.

$$n=\frac{e^x}{\ln(x)}$$ $$n\ln(x)-e^x=0$$

We see, your equation can be rearranged to an irreducible polynomial equation of more than one algebraically independent monomials ($\ln(x),e^x$). We therefore don't know how to rearrange the equation for $x$ by applying only finite numbers of elementary functions (elementary operations) we can read from the equation.
I don't know if the equation has solutions in the elementary numbers.

$$n=\frac{e^x}{\ln(x)}$$ $x\to e^z$:
for real $x,z$: $$n=\frac{e^{e^z}}{z}$$ $$ze^{-e^z}=\frac{1}{n}$$

We see, your equation cannot be solved in terms of Lambert W, but in terms of Hyper Lambert W:

$$G(-1;z):=ze^{-e^z}$$ $$G(-1;z)=\frac{1}{n}$$ $$z=HW(-1;\frac{1}{n})$$ $$x=e^{HW(-1;\frac{1}{n})}$$

So we have a closed form for $x$, and the representations of Hyper Lambert W give some hints for calculating $x$.

Galidakis, I. N.: On solving the p-th complex auxiliary equation $f^{(p)}(z)=z$. Complex Variables 50 (2005) (13) 977-997

Galidakis, I. N.: On some applications of the generalized hyper-Lambert functions. Complex Variables and Elliptic Equations 52 (2007) (12) 1101-1119

IV_
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Concerning the largest root (when it exists), you can have a quite good approximation of it after one single iteration of Halley method starting with $x_0=\log(n)$ writing the function as $$f(x)=x-\log(\log(x))-\log(n)$$ Using $$L_1=\log(n) \qquad\qquad L_2=\log(L_1)\qquad\qquad L_3=\log(L_2)$$ we obtain $$x_1=L_1+\frac{2 L_1 L_2L_3(L_1 L_2-1) }{2 (L_1 L_2-1)^2+(L_2+1) L_3}$$ Using this estimate, one single iteration of Newton method would give almost the solution $$x_2=x_1+\frac{x_1 \log (x_1) (\log (n)-x_1+\log (\log (x_1)))}{x_1 \log (x_1)-1}$$

Using $n=10^k$, some results $$\left( \begin{array}{cccc} k & x_1 & x_2 & \text{solution} \\ 2 & 5.09162713306291 & 5.09236408502652 & 5.09236407321316 \\ 3 & 7.61551194849834 & 7.61590897000061 & 7.61590896893251 \\ 4 & 10.0461676672324 & 10.0463804200754 & 10.0463804199298 \\ 5 & 12.4373422964443 & 12.4374674557571 & 12.4374674557281 \\ 6 & 14.8068649302967 & 14.8069444919917 & 14.8069444919841 \\ 7 & 17.1628132076291 & 17.1628668931673 & 17.1628668931649 \\ 8 & 19.5095090608892 & 19.5095470059521 & 19.5095470059513 \\ 9 & 21.8495239633877 & 21.8495517890709 & 21.8495517890705 \\ 10 & 24.1845060486885 & 24.1845270719237 & 24.1845270719235 \\ 11 & 26.5155717024025 & 26.5155879824867 & 26.5155879824866 \\ 12 & 28.8435102122394 & 28.8435230826814 & 28.8435230826813 \\ 13 & 31.1688990269469 & 31.1689093823764 & 31.1689093823764 \\ 14 & 33.4921726027437 & 33.4921810617013 & 33.4921810617012 \\ 15 & 35.8136655305610 & 35.8136725319265 & 35.8136725319265 \end{array} \right)$$

We could even do better using the first iterate of Householder method for the generation of $x_1$ but the formula would become quite nasty.

Claude Leibovici
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Mellin Barnes Like Integral Representation:

$\def\M{\operatorname M} \def\B{\operatorname B}$ We apply the Mellin Inversion theorem on the exponentially decreasing branch of $f^{-1}(x)=h(x)$:

enter image description here

$$f(x)=-e^{-x}\ln(x)\implies h(x)=\M_x^{-1}(\M_s(h(t))$$

The Mellin transform uses $e^y$’s Maclaurin series and integration by parts:

$$\M_s(h(t))=\int_0^\infty h(t)t^{s-1}dt=\frac{(-1)^s}{s!}\int_0^1 e^{-st}\ln^s(t)dt=\Gamma(s)\sum_{n=0}^\infty\frac{(-s)^n}{(n+1)^{s+1}n!}$$

Therefore:

$$\bbox[5px,border: 3.5px solid blue]{f(x)=\frac{e^x}{\ln(x)}\implies f^{-1}(x)=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}(-x)^s f_{-s-1}(-s)ds,f_a(x)=\sum_{n=0}^\infty \frac{(n+1)^ax^n}{n!}}$$

The integral of the summand nor the sum itself has a closed form yet, but $f_a(x)$ relates to the Bell polynomial $\B_n(x)$ for $a\in\Bbb N$. The Mathematica code InverseMellinTransform[Gamma[s]*Sum[(-s)^n/((n+1)^(s+1)n!),{n,0,b}],s,-1/x] for large $b$ and given $x$ verifies the result:

enter image description here

Series Solution:

Using the Bell polynomial and Lagrange reversion gives this series for the same branch of the inverse function:

$$\bbox[5px,border: 3.5px solid blue]{f(x)=\frac{e^x}{\ln(x)}\implies f^{-1}(x)=\sum_{n=1}\frac{e^\frac nx}{n!}\B_{n-1}\left(\frac nx\right)}$$

shown here

Тyma Gaidash
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  • Note: one gets the following Ramanujan master theorem series where one of the series likely has a closed as a Bell polynomial. The result seems to be the same as that of Lagrangr reversion – Тyma Gaidash Feb 21 '24 at 21:25