# Finding a real closed-form solution to a tricky transcendental equation

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?

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

$$\left(c+\frac{1}{x}\right)^x=x \qquad \implies \qquad c=x^{\frac{1}{x}}-\frac{1}{x}$$ $$\frac {dc}{dx}=\frac{x^{\frac{1}{x}} (1-\log (x))+1}{x^2}$$ 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 $$1.74\times 10^{-8}$$.

Edit

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:

$$x^{1/x}(1-\log(x))=-1$$

or:

$$x^{1/x}(\log(x)-1)=1$$

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 $$c$$