One of my friends posed the following question with a prize of free bubble tea for anyone who could find a closed-form solution to the problem (not necessarily in terms of elementary functions):

What is the maximum value of $c$ such that $(c + \frac{1}{x})^x=x$ has real solutions?

I've been easily able to whittle this problem down to solving the equation $r^r+\text{ln }r+1=0$ (where the solution to the problem is $c=r^r-r$) but from there I don't really know how to proceed - I've already tried a bit of variable shuffling using Lambert's W function, but with no luck. I've also tried to solve similar equations (like $r^r + \text{ln }r=0$) first to see if they could give me further insight to the problem - still no dice. Are there any other tools or approaches that could yield fruit? Or perhaps is there a way to show this problem isn't solvable neatly (for some definition of neatly) at all?

  • $\begingroup$ what is your $r$? $\endgroup$
    – MathFail
    Jul 25, 2023 at 3:11
  • $\begingroup$ Determine intervals where $r^r+\ln r +1$ is increasing or decreasing. $\endgroup$
    – GEdgar
    Jul 25, 2023 at 5:35
  • $\begingroup$ I believe most closed possible form is $c=r^{-r}-r$, where $r$ is real root of $r^r+\ln r+1=0$ $\endgroup$ Jul 25, 2023 at 8:36

3 Answers 3


$$\left(c+\frac{1}{x}\right)^x=x \qquad \implies \qquad c=x^{\frac{1}{x}}-\frac{1}{x}$$ $$\frac {dc}{dx}=\frac{x^{\frac{1}{x}} (1-\log (x))+1}{x^2}$$ which is, in slightly different words, what you properly wrote.

So, the problem is to find the zero of function $$f(x)=x^{\frac{1}{x}} (1-\log (x))+1$$ Using inspection, we know that the solution is somewhere between $e^{3/2}$ and $e^2$ since $$f\left(e^{3/2}\right)= 1-\frac{1}{2} e^{\frac{3}{2 e^{3/2}}}=0.301$$ $$ f\left(e^{2}\right)=1-e^{\frac{2}{e^2}}=-0.311$$

Between these two values, the function is close to linearity. So, as a guess, use the midpoint $$x_0=\frac{1}{2} \left(e^{3/2}+e^2\right)$$ and perform one single iteration of any Newton-like method of order $n$ to obtain a nasty but fully explicit expression for $x_1^{(n)}$.

Converting the results to decimals, the results are (respectively for Newton and Halley methods) $$x_1^{(2)}=\color{red}{5.6}6977 \qquad \text{and} \qquad x_1^{(3)}= \color{red}{5.6773}3$$ while the "exact" solution is $x=5.67737$.

The number $$x=5.6773739131926818822733208953370077229026966798039\cdots$$ is not identified by inverse symbolic calculators but it is surprizingly close to the largest root of the quartic equation $$3 x^4+20 x^3+100 x^2-10000=0$$ which can be solved with radicals.

The absolute difference is $3.04\times 10^{-7}$.

Since we know $x$, just compute $c$.

Using the $ISC$ a good approximation is $$c_{\text{max}} \sim e^{-\sqrt{2}} \left(\tan ^{-1}\left(\frac{1}{2}\right)+\Gamma \left(\frac{5}{24}\right)\right)$$ which is an absolute error of $2.37\times 10^{-8}$.

Another approximation is $$c_{\text{max}} \sim 2 \,\Omega -\frac 1{100}\left(\psi\left(\frac{3}{16}\right)-\psi \left(\frac{7}{12}\right)+\psi \left(\frac{13}{14}\right)\right)$$which is an absolute error of $1.74\times 10^{-8}$.


Pushing the inspection, a very good guess is $$x_0=e^{\sqrt{3}}\qquad \implies \qquad f(x_0)=0.0056$$ which gives $$x_1^{(2)}=\color{red}{5.6773}01 \qquad \text{and} \qquad x_1^{(3)}= \color{red}{5.67737}4$$.


Let $x=e^{e^t+1}$ and apply Lagrange reversion:

$$\sqrt[x]x(\ln(x)-1)=1\iff -(e^t+1)e^{-(e^t+1)}=t\implies t=\sum_{n=1}^\infty\frac{(-1)^n}{n!}\left.\frac{d^{n-1}}{dw^{n-1}}(e^w+1)^ne^{-n(e^w+1)}\right|_0$$

If we solved for $x$ as just a series, then we would evaluate $\left.\frac{d^{n-1}}{dw^{n-1}}e^w(e^w+1)^ne^{(1-n)(e^w+1)}\right|_0$. To evaluate it as a double sum, one would only think to use $e^y$’s Maclaurin series, but that series as if diverges for larger $n$. Continuing, apply Stirling S2, general Leibniz rule, and factorial power $n^{(m)}$

$$\left.\frac{d^{n-1}}{dw^{n-1}}(e^w+1)^ne^{-n(e^w+1)}\right|_0=\sum_{k=0}^{n-1}s_{n-1}^{(k)}\left.\frac{d^ke^{-nw}(w+1)^n}{dw^k}\right|_1=\sum_{k=0}^{n-1}\sum_{m=0}^k\binom km s_{n-1}^{(k)}\frac{n^{(m)}(-1)^{k+m+n}n^{k-m}2^{n-m}}{e^{2n}n!}$$

Summing over $n$ gives the Tricomi confluent hypergeometric function:

$$\bbox[3px,border:3.5px ridge #8aecff ]{\max\left(\left(\frac1x+c\right)^x=x\right):x=\exp\left(1+\exp\left(\sum_{n=1}^\infty\sum_{k=0}^n\frac{s_{n-1}^{(k)}(-2)^{n-k}}{e^{2n}n!}\operatorname U(-k,n-k+1,2n)\right)\right)}$$

shown here


This is not an answer but too long to put as a comment. Starting with Claude's observation that we can instead try to solve:

$\frac{x^{1/x}(1-\log(x))+1}{x^2} = 0 $

note we can rewrite as:




taking the log of both sides:

$\frac{1}{x}\log(x) + \log(\log(x)-1) = 0$

substituting $y = \log(x)$:

$y e^{-y} + \log(y -1) =0 $

And this problem seems a little easier to approach. I still do not know how to find the root however.

Using Claude's approximation $x_0 \approx e^{\sqrt{3}}$ (or $y_0 \approx \sqrt{3}$) we can also do an iteration of Newton's method to get new approximations:

$y^{(1)}_0 = \frac{-3 \sqrt{3}+3+e^{\sqrt{3}} \left(\sqrt{3}-\left(\sqrt{3}-1\right) \log \left(\sqrt{3}-1\right)\right)}{2 \sqrt{3}+e^{\sqrt{3}}-4} ~~\mathrm{ and }~~ x^{(1)}_0 = e^{y^{(1)}_0} \approx \color{red}{5.677}29$

$y^{(2)}_0$ is too ugly to be worth showing here but it yields a very good approximation:

$x^{(2)}_0 = e^{y^{(2)}_0} \approx \color{red}{5.67737391}2$

I know I didn't really add anything new compared to Claude's answer, but I think looking at $y e^{-y} + \log(y -1) =0 $ instead may be easier for finding (if possible) an exact solution for $x_0$ and then plugging into the original equation to find $c$


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