# For what real value $k$ does the equation $e^{-kx}+e^{-x} = x$ have only one real root?

I am working on this equation, and I think I need to consider Rolle's theorem. According to a teacher, I need to use IMV as well. Any ideas?

$e^{-kx}+e^{-x} = x$

The value of $k$ is that, wherein the exponential $f(x) = e^{-kx}+e^{-x}$ is tangent to $g(x) = x$.

• For every $k \geq 0$, the graphs of functions $f(x) = e^{-kx} + e^{-x}$ and $g(x) = x$ intersect in one and only one point. – jgsmath Jul 2 '17 at 20:02
• Yes. :) How would one do it with $k < 0$? It is double-valued. – mov0021 Jul 3 '17 at 1:13

Take the derivatives and set equal...

$$-ke^{-kx}-e^{-x}=f'(x)=g'(x)=1$$

$$-ke^{-kx}=1+e^{-x}$$

Clearly $k\ge0$ is not possible, so we look at $k=-n<0$

$$ne^{nx}=1+e^{-x}$$

$$u=e^x\implies nu^{n+1}=1+u$$

When $n$ is even, then there are two solutions for $u$ and when $n$ is odd, there is only one solution.

(Prove with intermediate value theorem)

Then check if $f(x)=g(x)$ at those points.

• The OP did not state that $k$ has to be integral. – John Bentin Jul 2 '17 at 20:46

If $k>0$, left side is monotonously decreasing to 0 and right side increasing, so it clearly always has 1 and only 1 real root. You can use Rolle's theorem for this case to prove it rigorously.

For $k<0$ left side is a convex function and it will cross $y=x$ either never (huge negative $k$), twice ($k$ just below zero), or (in a single $k$ inbetween when it has a double root). If it's a double root, it matches both in value and derivative. So the critical value of $k$ satisfies

$$e^{-kx}+e^{-x}=x$$ $$-ke^{-kx}-e^{-x}=1$$

which is two equations for variables $k$ and $x$.

Plotting a few cases for $k$, I fond the critical value to be somewhere around $-0.36$ which naturally gets me thinking of $k=-e^{-1}$ but isn't quite correct. In fact, the pair of equations is transcendental and has no reason of having a closed form solution.

EDIT:

I don't think this can be expressed only with Lambert's function. Here's my try. Setting $n=-k$, above can be converted to

$$e^{nx}=\frac{1+x}{1+n}=u$$ $$e^{-x}=\frac{nx-1}{n+1}=v$$ where $x=u+v$. The last one can be converted to $$ve^v=e^{-u}\rightarrow v=W(e^{-u})$$ I hoped that substitutio to $u$ and $v$ variables would fully separate the variables and enable me to solve for $u$ and $v$ separately, but the $nx=(u+v)(1+v)/u$ can't be nicely simplified to something that would lead to a solvable equation. There may be better substitutions but I don't think the system has enough symmetry for this to be successful.

This process, however, does lead to a nice iterative scheme which relates to the guessed approximations. The iteration begins with initial condition:

$$u_0=e, \quad v_0=0$$ These are consistent with the guess $n\approx e^{-1}$, $x\approx 1/n \approx e$, $nx\approx 1$.

The iterative step is:

$$u_{k+1}=e^{(1+v_k)(u_k+v_k)/u_k},\quad v_{k+1}=W(e^{-u_{k+1}})$$

This converges fairly quickly to $u=2.915444361$, $v=0.05146225679$, and therefore $x=2.96690661781$ and $n=0.36065248606699$.

• Yes, my friend. I love trascendental. Something around -0.360524. I still cannot find it exactly. – mov0021 Jul 2 '17 at 20:17
• You won't "find" it, unless you mean numerically or in terms of an infinite recursion. It might involve Lambert's function, but even that's buried in another layer of equations. – orion Jul 2 '17 at 20:24
• Why cannot I find it? Yes. Indeed it requires W-function, but that should be not as chaotic as numerical methods. In fact, the homework is to produce an analytical result which puts numerical methods aside. – mov0021 Jul 2 '17 at 21:33
• W-function is numerical function, it just got a name because it's used a lot (on the inside, it's doing one of the iterative procedures, or maybe a rational approximation). But yes, I know what you want, I'll be playing with it and see if I can express this in terms of W. – orion Jul 3 '17 at 6:30