How to solve $x \geq \frac{y}{z-\ln{x}}$ for positive variables?

Question

How can you solve $x \geq \frac{y}{z-\ln{x}}$ for $x$ when the variables are real positive values? I am only really interested in the case where the values are large and $z > \ln x$.

How can one find a closed form solution for $x$? I am ok with bounds that are out by constant factors.

Maple does not give an answer. For the equality $x = \frac{y}{z-\ln{x}}$ it gives
$$x = -\frac{y}{\text{LambertW} ( -y e^{-z} )}.$$

I don't really understand this Maple solution as the real values $z = 2\ln{x}$ and $y = x\ln x$ solve the original equation and the denominator in the Maple solution seems to be imaginary.

Is this Maple solution to the equality correct?

As a note, I believe that $\text{LambertW(n)} \sim \ln(n)$ and also $ \ln{n} \geq \text{LambertW(n)} >\ln/2$ if $n \geq e$ so this might be a useful way to get a bounds in terms of $\ln$.

Answer 1

If you are given $y,z$ and looking for $x$ you won't find it without the Lambert W function or a numeric approach. A numeric approach will work easily, though. You can either use a one-dimensional root finder such as in chapter 9 of Numerical Recipes or any numerical analysis text, or you can use iteration. If $x$ is large (and $10$ is getting large, $100$ certainly is), $x$ is much greater than $\ln x$. Start with $x_0= \frac yz$ and iterate $x_{i+1}=\frac y{z-\ln x_i}$. It should converge very quickly. Then if $x_c$ is the convergent, your solution is $x_c \le x\lt \exp(x)$

Answer 2

Since $z > \ln x$ and all variables are positive we have

$$ \begin{align} x \geq \frac{y}{z - \ln x} \quad &\Longleftrightarrow \quad \ln x - z \leq - \frac{y}{x} \\ &\Longleftrightarrow \quad x e^{-z} \leq e^{-y/x} \\ &\Longleftrightarrow \quad -ye^{-z} \geq - \frac{y}{x} e^{-y/x}. \end{align} $$

For the moment let $w = -y/x$, so that the above inequality is $-y e^{-z} \geq w e^w$. The quantity $w e^w$ has a minimum at $w=-1$ with height $-1/e$, so there are no solutions to your inequality if $-y e^{-z} < -1/e$ or, equivalently, if $y > e^{z-1}$.

Assuming solutions exist, we can solve your inequality using the two real-valued branches of the Lambert $W$ function, $W_0$ and $W_{-1}$. We get

$$ W_{-1}(-ye^{-z}) \leq -\frac{y}{x} \leq W_0(-ye^{-z}) $$

or, equivalently,

$$ - \frac{y}{W_{-1}(-ye^{-z})} \leq x \leq - \frac{y}{W_{0}(-ye^{-z})} $$

It is known that $W_{-1}(x) \leq \ln(-x) - \ln(-\ln(-x))$ for $-1/e \leq x < 0$, which we can use to find a lower bound on $x$. We find that the lower bound $x \geq -y/W(-ye^{-z})$ is satisfied when

$$ \frac{y}{z-\ln y + \ln(z-\ln y)} \leq x. $$

By referring to the asymptotic formula for $W_{-1}$ we can see that the absolute error of this approximation decreases to $0$ as $y e^{-z} \to 0$.

Bounding $x$ above is not as easy. By the power series expansion of $W_0(z)$ about $z=0$ we know that

$$ \begin{align} -\frac{1}{W_0(-z)} &= \frac{1}{z} - 1 - \frac{1}{2}z-\frac{2}{3}z^2-\frac{9}{8}z^3-\frac{32}{15}z^4-\frac{625}{144}z^5 - \cdots \\ &= \frac{1}{z} - \sum_{n=0}^{\infty} {c_n} z^n, \end{align} $$

which is convergent for $0 < z < 1/e$ and where every coefficient $c_n$ is negative. To get a lower bound for this we may truncate this series at any point and conclude that

$$ -\frac{1}{W_0(-z)} \geq \frac{1}{z} - \left(\sum_{n=0}^{N} {c_n} z^n\right) - 2 c_{N+1} z^{N+1} $$

for all $z > 0$ small enough. (Note the $2$ here is almost arbitrary; the factor only needs to be $>1$ and could probably be chosen to maximize the interval where the bound holds.) For example,

$$ -\frac{1}{W_0(-z)} \geq \frac{1}{z} - 1 - z $$

for $0 < z \leq 0.307$, and thus

$$ - \frac{y}{W_0(-ye^{-z})} \geq e^z - y - y^2 e^{-z} $$

for $0 \leq y \leq 0.307 e^z$. The absolute error on this lower bound is $O(y^3 e^{-2z})$.

We conclude that

For the inequality

$$ x \geq \frac{y}{z - \ln x} $$

to be satisfied it is sufficient to have

$$ \frac{y}{z-\ln y + \ln(z-\ln y)} \leq x \leq e^z - y - y^2 e^{-z} $$

as long as $0 \leq y \leq 0.307 e^z$.