Is the function $y=xe^x$ invertible?

Question

I'm wondering if the equation $re^r=se^s$ has any answer. If there is any answer,and $r=-1+v,s=-1-v$ in which $v$ is a positive real number,what can we say about $v$?

Thank you in advance.

Answer 1

Let $r=-1+v$ and $s=-1-v$. You are looking for zeros of the function: $$ f(v) = (-1+v)e^{-1+v} - (-1-v)e^{-1-v}. $$ Observe that one solution is $v=0$ which corresponds to $r=s=-1$. Compute the derivative $$ f'(v) = ve^{-1-v} (-1+e^{2 v}) $$ and notice that it is never negative and it is zero only for $v=0$. Hence $f$ is strictly increasing. This means $f$ is injective and there cannot be other solutions. So you can say that $v=0$.

Answer 2

The derivative is $e^x+xe^x$ which vanishes at $-1$. The second derivative is $2e^x+xe^x$, which is $2e^{-1}-e^{-1}=e^{-1}$ at $-1$. The limit at $-\infty$ is $0$ and the limit at $+\infty$ is $+\infty$, both of which are bigger than $-e^{-1}$. From this you can read off everything there is to say about the number of solutions to $xe^x=a$: the solution is unique for $a \geq 0$ and $a=-e^{-1}$, there are two solutions for $a \in (-e^{-1},0)$, and there are no solutions for $a<-e^{-1}$.

As an aside, the solutions for $a \geq 0$ create an inverse for (a restriction of) $xe^x$. This function (or rather its analytic continuation) is called the Lambert W function.

Answer 3

Summary: For $re^r=se^s$ with $r=-1+v, \,s=-1-v$ there is only one solution: $v=0$.
Thus $r = s = -1$, and $re^r=se^s = -e^{-1}$.

It's not possible to write $s$ as an explicit function of $r$ (or vice versa) using elementary functions. Clearly, there are infinite trivial solutions, with $r = s$. And we can generate other solution pairs by parametrization:

Let $r = st$, with $t > 1$

$$\begin{align} st e^{st} & = se^s\\ t e^{st} & = e^s\\ e^{st} e^{-s} & = t^{-1}\\ e^{s(t-1)} & = t^{-1}\\ e^s & = t^{\frac{1}{1-t}}\\ s & = \frac{\log t}{1-t}\\ r & = st = \frac{t\log t}{1-t}\\ \end{align}$$

But we want $r=-1+v, \,s=-1-v$, so $r+s = -2$, i.e., $\left(\frac{1 + t}{1-t}\right)\log t = -2$.

Here's a portion of the graph of $y=\left(\frac{1+x}{1-x}\right)\log x$, courtesy of Wolfram Alpha.

It appears that $y=-2$ when $x=1$.

We can't directly evaluate $\left(\frac{1 + t}{1-t}\right)\log t$ when $t=1$, since it's indeterminate, but we can evaluate it using L'Hôpital's rule:

Let $u = (1 + t)\log t$, $v = 1 - t$
$u' = \frac{1 + t}{t} + \log t$, $v' = -1$
At $t=1$, $\frac{u'}{v'} = \frac{2/1}{-1} = -2$

So $t=1$, and since $r = st$, $r = s$. Hence $v = 0$ and $r = s = -1$, as stated above.

Strictly, we should show that $y=\left(\frac{1+x}{1-x}\right)\log x$ has a maximum at (1, -2), and is hence a unique solution. This gets a little messy, due to that indeterminate form.

$$\begin{align} y & =\left(\frac{1+x}{1-x}\right)\log x\\ \frac{dy}{dx} & = \left(\frac{1+x}{1-x}\right)x^{-1} + \left(\frac{(1-x) + (1+x)}{(1-x)^2}\right)\log x\\ & = \frac{(1+x)(1-x) x^{-1}}{(1-x)^2} + \frac{2\log x}{(1-x)^2}\\ & = \frac{(1-x^2) x^{-1} + 2\log x}{(1-x)^2}\\ \frac{dy}{dx} & = \frac{x^{-1} - x + 2\log x}{1 - 2x + x^2} \end{align}$$

At $x=1$, $\frac{dy}{dx}$ is indeterminate. Applying L'Hôpital's rule, we get

$$\frac{-x^{-2} - 1 + 2x^{-1}}{-2 + 2x}$$

which is still indeterminate at $x=1$. Applying L'Hôpital's rule again:

$$\frac{2x^{-3} - 2x^{-2}}{2}$$

which is zero at $x=1$. Thus (1, -2) is a stationary point of the function, and from the graph we can see that it's a maximum, so I won't bother messing around with second derivatives. :)

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FWIW, we can transform $re^r=se^s$ into other interesting forms.

Let $r=\log(R)$, $s=\log(S)$.
Then $re^r=se^s \implies$

$$\begin{align} R\log(R) & = S\log(S)\\ R^R & = S^S\\ R^{\frac{1}{S}} & = S^{\frac{1}{R}}\\ \frac{1}{R}^{\frac{1}{S}} & = \frac{1}{S}^{\frac{1}{R}}\\ \text{Let } x & = \frac{1}{R}, y = \frac{1}{S}\\ x^y & = y^x \end{align}$$

Note that when $r = s = -1$, $x = y = e$.

Also, $r + s = -2 \implies RS = e^{-2} \implies xy = e^2$

Here's the graph of $x^y = y^x$. The hyperbola-like arc is crossed by the line $y = x$ at $y = x = e$, and in the neigbourhood of $y = x = e$ that arc approaches the hyperbola $xy = e^2$, as I mentioned in this recent answer and its comments.

In that answer, I developed a parametrization for $x^y = y^x$, which is (obviously) closely related to the one in this answer. In the variation mentioned in the comment we have

$$x=\left(1+\frac{1}{n}\right)^n, \, y=\left(1+\frac{1}{n}\right)^{n+1}$$

That formula for $x$ should be familiar from the well-known limit definition of $e$: $$e = \lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n$$

Clearly $y$ also approaches $e$ as $n \rightarrow \infty$ and hence $xy \rightarrow e^2$ as $n \rightarrow \infty$

It can be shown that as $n \rightarrow \infty$ $x$ approaches $e$ from below, and $y$ and $\sqrt{xy}$ both approach $e$ from above. The proof will be left as an exercise for the reader :) , but here's a short empirical demonstration.

n:          x          y   sqrt(xy)
1: 2.00000000 4.00000000 2.82842712
2: 2.25000000 3.37500000 2.75567596
3: 2.37037037 3.16049383 2.73706794
4: 2.44140625 3.05175781 2.72957517
5: 2.48832000 2.98598400 2.72581799
6: 2.52162637 2.94189743 2.72366778
7: 2.54649970 2.91028537 2.72232269
8: 2.56578451 2.88650758 2.72142544
9: 2.58117479 2.86797199 2.72079713

And so we see that there are no other solutions for the question in the OP, since there's only a single solution for $xy = e^2$

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Notice how $\sqrt{xy}$ seems to give better approximations to $e$ than $x$ or $y$ do. But we can do even better with a slight modification.

Let $m = n - \frac{1}{2} + \frac{12}{n}$. Then compute $\left(1+\frac{1}{m}\right)^n$

  n:       (1 + 1/m)^n
  1: 2.714285714285714
  2: 2.718042366691015
  4: 2.718267026012865
  8: 2.718280905875519
 16: 2.718281770837749
 32: 2.718281824858342
 64: 2.718281828234011
128: 2.718281828444980
256: 2.718281828458166
512: 2.718281828458990

Answer 4

A straightforward solution is

$$ r=s $$

for whatever parametrization. Other solution involves Lambert function $W$

$$ r + \log r = s + \log s$$