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From deeper discussion of this question I have the pair of equations $$ t = \exp(x - u) \\ u = \exp(x - t) $$ Rearranging $$ t/u = \exp(t)/\exp(u) \qquad \text{ or }\\ u/\exp(u) = t/\exp(t) $$

I'm looking at $0 \le x \le 1$ if this is simpler. My hypothese is that $t=u$ is required.
Trying with arithmetic or geometric mean of $u$ and $t$ didn't lead easily to a solution, also the use of the taylor-series for $\exp()$ didn't give me the idea.

I think however it must be somehow easy to be proved, but after a time fiddling with it the penny still didn't drop...

Q: does the hypothese hold that for all $x$ in the interval $t=u$ is required?

Gottfried Helms
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2 Answers2

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Rearranging, you get: $$ \tag{1} te^u = e^x = ue^t $$ and from there $$ \tag{2} \frac{t}{e^t} = \frac{u}{e^u} $$ The function $t\mapsto t/e^t$ increases from $t=0$ to $t=1$, but after that it starts falling again -- so you can find $t\ne u$ that satisfies $(2)$ quite fine, and from there use $(1)$ to compute $x=\log(t)+u$.

The only question is whether $x$ will end up in your range from $0$ to $1$. Unfortunately it never does. I don't have a slick theoretical proof of this, but by numerical experiment, as $t$ increases from $1$ to $e$ (after which point there is clearly no hope because the $\log t$ term alone will be too large), the required $x$ seems to increase monotonically from $1$ to about $1.22$.

(For $t\ge 1$ solving for $u$ gives $u=-W(-t/e^t)$, so we can get Wolfram Alpha to plot $x$ as a function of $t$).

So there is no solution with $t\ne u$ given your constraints on $x$ -- but for every $x>1$ there is one.

hmakholm left over Monica
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  • Hmm, this seems to be rather complete now. I'll try to apply this problem of when $x \gt 1$ to the basic question which I've linked to. – Gottfried Helms Dec 24 '18 at 01:37
  • Next idea: I get that $u=W(\exp(x+a))$ and $t=W(\exp(x-a))$ . It has to be shown that $a=0$ is required when $x \in 0..1$ . Perhaps this can make it smoother, but yet I do not see how.... – Gottfried Helms Dec 24 '18 at 08:28
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I've got possibly an ansatz, but get stuck at the end.

$$ \begin{array} {rl|rl|l} t &= \exp(x-u) & u &= \exp(x-t) & \implies \\ t \exp(u) &= \exp(x) & u \exp(t) &= \exp(x) & \\ \end{array} $$ so we have $$ t \exp(u) = u \exp(t) \qquad \gt 0 $$ See a contourplot for $\log(t)+u = \log(u)+t$ drawn by W/A: image

Assume now, that $u=t+d$ then we get $$ t \exp(t) \exp(d) = (t+d)\exp(t) \\ t \exp(d) = t+d \\ t (1+d+d^2/2+ \cdots) = t+d \qquad \text{ we know $t \gt 0$ so we can cancel}\\ (1+d+d^2/2+ \cdots) = 1+d/t \\ d+d^2/2+ \cdots = d/t \\ $$ Hmm. If $t \ge 1$ we have our contradiction and $d=0$ is required. But if $t \lt d$ I'm again stuck....

Gottfried Helms
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