Finding the Inverse of a Function that Involves the Natural Logarithm

Question

Let $v(x)=e^\gamma x\ln\left(\ln\left(e^\gamma x\right)\right)+\ln\left(e^\gamma x\right)$, where $\gamma$ is the Euler–Mascheroni constant ($0.5772$...) and let $v_1(x)=\frac{v(x)}{x \ln \ln x}-e^\gamma$ Let $b$ be any positive number, what is the general formula to find the $x_1$ so that $v_1(x_1)=b$?

EDIT: I realized that another (maybe easier) way of saying what I want is I want the inverse of $v_1(x)$, so that $x=\frac{v(y)}{y \ln \ln y}-e^\gamma$. I want to solve for $y$.

Answer 1

As Ross Millikan answered and commented, there is no hope for a closed form expression of $x(b)$ and numerical methods would be required.

In my opinion, whatever you would do, I think that it will be better to consider that you look for the zero of $$f(x)=\frac 1 {v_1(x)}-\frac 1b$$ since $\frac 1 {v_1(x)}$ is much more linear than ${v_1(x)}$.

You could have a quite good approximation of the solution at the price of a quadrating equation in $(x-e)$ building the $[2,2]$ Padé approximant of $\frac 1 {v_1(x)}$ around $x=e$. The analytical formula is quite ugly but, numerically, it is $$\frac 1 {v_1(x)}\sim \frac {\frac{263 }{995}(x-e)+\frac{82 }{1651}(x-e)^2 } {1+\frac{311 }{475}(x-e)+\frac{8}{683} (x-e)^2}$$

Trying with $b=12.345$, this would give an estimate equal to $3.03780$ while the exact solution (obtained using Newton method) would be ... the same !

Trying with $b=1.2345$, this would give an estimate equal to $6.97428$ while the exact solution (obtained using Newton method) would be $7.00838$ !

Answer 2

We need $x \gt e$ for $v_1(x)$ to be positive. $v_1(x)$ gets large and positive as $x$ approaches $e$ from above because the $\log(\log(x))$ term in the denominator becomes small. A plot is below with $\gamma$ approximated. You are in for numeric root finding, so the challenge is to find a good starting value. Let $x=e^z$, with $z \gt 1$, then $$v_1(e^z)=\frac {e^\gamma x\ln\left(\ln\left(e^\gamma x\right)\right)+\ln\left(e^\gamma x\right)}{x \log \log x}-e^\gamma\\ =\frac{e^\gamma e^z \ln (\gamma+z)+(\gamma +z)}{e^z \log z}-e^\gamma$$ I would pick some value for $z$ like $1.1$ or $1.5$ for all the entries except the $\log z$ in the denominator and compute $z$ from that, then feed it to my favorite root finder.